NDA Mathematics 21 Apr 2019 Official Paper (Previous Year Paper)

Concept:

If a₁, a₂, …., a_n is a GP then the general term is given by: a_n = a × r^{n - 1} where a is the 1^st term and r is the common ratio.

Calculation:

Given: 25, -125, 625, -3125, …….

Here, first term a = 25 and common ratio r = -5.

As we know that, if a₁, a₂, …., a_n is a GP then the general term is given by: a_n = a × r^{n - 1} where a is the first term and r is the common ratio.

⇒ The general term is : a_n = 25 × (-5)^{n – 1} = (-1)^{n – 1} × 5^{n + 1}

Concept:

For any two non-empty sets A and B, a relation R from A to B is a subset of the cartesian product A X B.

Reflexive:

Let R be a relation on a non-empty set A, if every element of A is related to itself then R is said to be a reflexive relation.

Thus, R is reflexive if and only if (a, a) ∈ R, ∀ a ∈ A.

Symmetric:

Let R be a relation on a non-empty set A, then the relation R is said to be symmetric relation if and only if (a, b) ∈ R ⇒ (b, a) ∈ R

Transitive:

Let R be a relation on a non-empty set A, then the relation R is said to be transitive relation if and only if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R

Equivalence:

Let R be a relation on a non-empty set A, then the relation R is said to be equivalence relation if R is reflexive, symmetric and transitive.

Calculation:

Given: X = {1, 2, 3, 4} and R is a relation on X given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2), (1, 3), (3, 1)}

Reflexive:

The given relation R on X is not reflexive ∵ (a, a) ∉ R, ∀ a ∈ X i.e (4, 4) ∉ R.

Symmetric:

The given relation R on X is symmetric ∵ for all (a, b) ∈ R ⇒ (b, a) ∈ R 

Transitive:

The given relation R on X is transitive ∵ for all (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R

Concept:

For any two non-empty sets A and B, a relation R from A to B is a subset of the cartesian product A X B.

Reflexive:

Let R be a relation on a non-empty set A, if every element of A is related to itself then R is said to be a reflexive relation.

Thus, R is reflexive if and only if (a, a) ∈ R, ∀ a ∈ A.

Symmetric:

Let R be a relation on a non-empty set A, then the relation R is said to be symmetric relation if and only if (a, b) ∈ R ⇒ (b, a) ∈ R ∀ a, b ∈ A.

Transitive:

Let R be a relation on a non-empty set A, then the relation R is said to be transitive relation if and only if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R, ∀ a, b, c ∈ A.

Equivalence:

Let R be a relation on a non-empty set A, then the relation R is said to be equivalence relation if R is reflexive, symmetric and transitive.

Calculation:

Given: R is a relation on N and is defined as xRy ⇒ x² – 4xy + 3y² = 0.

⇒ x² – 4xy + 3y² = 0

⇒ (x - y) (x – 3y) = 0     ...1)

Reflexive:

The given relation R is reflexive ∵ if x = a and y = a then (x = a, y = a) satisfies the equation (1) ⇒ (a, a) ∈ R, ∀ a ∈ N.

Hence, R is reflexive.

Symmetric:

∵ (3, 1) satisfies the equation (1) ⇒ (3, 1) ∈ R.

Let’s check whether (1, 3) ∈ R or not. If we substitute x = 1 and y = 3 in equation (1), we get

⇒ (x - y) (x – 3y) = 16 ≠ 0.

⇒ (1, 3) ∉ R.

Hence, the given relation R is not symmetric ∵ (3, 1) ∈ R but (1, 3) ∉ R.

Transitive:

As we can see that (9, 3) and (3, 1) satisfy the equation (1).

⇒ (9, 3) ∈ R and (3, 1) ∈ R.

Let’s check whether (9, 1) satisfies the equation (1) or not.

⇒ (x - y) (x – 3y) = (9 - 1) (9 - 3) = 48 ≠ 0.

⇒ (9, 1) ∉ R.

Hence, the given relation R is not transitive ∵ (9, 3) ∈ R and (3, 1) ∈ R but (9, 1) ∉ R.

Concept:

1, ω and ω² are called cube roots of unity where $\omega = \frac{{ - 1 + i\sqrt 3 }}{2};and;{\omega ^2} = \frac{{ - 1 - i\sqrt 3 }}{2}$

1 + ω + ω² = 0 and ω³ = 1.

Calculation:

Given: A = {x ϵ Z : x³ – 1 = 0} and B = {x ϵ Z: x² + x + 1= 0}

As we know that, the roots of the x³ – 1 = 0 are 1, ω and ω².

⇒ A = {1, ω, ω²}

As we know that, the roots of x² + x + 1= 0 are ω and ω².

⇒ B = {ω, ω²}

∴ A ∩ B = B

Alternate MethodThe roots of equation x3 – 1 = 0 are

\(\left{ {1, \frac{{-1 + \sqrt 3 i}}{2},\frac{{-1 - \sqrt 3 i}}{2}} \right}\)

And the roos of the equation x2 + x + 1 = 0 are

\(\left{ {\frac{{-1 + \sqrt 3 i}}{2},\frac{{-1 - \sqrt 3 i}}{2}} \right}\)

So, the common roots between A and B are, \(\left{ {\frac{{-1 + \sqrt 3 i}}{2},\frac{{-1 - \sqrt 3 i}}{2}} \right}\)

Therefore, A ∩ B = \(\left{ {\frac{{-1 + \sqrt 3 i}}{2},\frac{{-1 - \sqrt 3 i}}{2}} \right}\)

Concept:

A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

Calculation:

Let, us suppose the universal set U = A ∪ B.

Statement 1:

⇒ (A ∩ B) ∪ (A ∩ B̅) ∪ (A̅ ∩ B) = {A ∪ (A ∩ B̅)} ∩ {B ∪ (A ∩ B̅)} ∪ (A̅ ∩ B)

⇒ (A ∩ B) ∪ (A ∩ B̅) ∪ (A̅ ∩ B) = {(A ∪ A) ∩ (A ∪ B̅)} ∩ {(A ∪ B) ∩ (B ∪ B̅)} ∪ (A̅ ∩ B)

⇒ (A ∩ B) ∪ (A ∩ B̅) ∪ (A̅ ∩ B) = {A ∩ U} ∩ {U ∩ U} ∪ (A̅ ∩ B)

⇒ (A ∩ B) ∪ (A ∩ B̅) ∪ (A̅ ∩ B) = A ∪ (A̅ ∩ B) = (A ∪ A̅) ∩ (A ∪ B) = A ∪ B

Hence statement 1 is Correct.

Statement 2:

⇒ (A ∪ (A̅ ∩ B̅)) = (A ∪ A̅) ∩ (A ∪ B̅)

⇒ (A ∪ (A̅ ∩ B̅)) = U ∩ (B ∩ A̅) = A̅ ∩ B

Hence, statement 2 is wrong.

Concept:

1. A ∆ B is the symmetric difference that represents the objects that belong to A or B but not to their intersection.

• A Δ B = (A ∪ B) - (A ∩ B)

2. A ⊆ B subset A is a subset of B. set A is included in set B.
3. A ⋃ B, ⋃ is a union, that represents the objects that belong to set A or set B
4. A ⋂ B, ⋂ is an intersection, that represents the objects that belong to set A and set B
5. a ∈ A means x is an element of A, i.e. belongs to set membership

Calculation:

Given: A, B, C be subsets of X

Statement 1: A ⊂ C ⇒ (A ∩ B) ⊂ (C ∩ B), (A ∪ B) ⊂ (C ∪ B)

We know that,

If A ∪ B = A ∪ C and A ∩ B=A ∩ C, then B = C which gives 

⇒ A ⊂ C 

⇒ A ⊂ C ⇒ (A ∩ B) ⊂ (C ∩ B), (A ∪ B) ⊂ (C ∪ B)

Hence statement 1 is true.

Statement 2: (A ∩ B) ⊂ (C ∩ B) for all sets B ⇒ A ⊂ C

Let A = {1, 2, 3}, B = {3, 4, 5} and C = {1, 3, 6, 7, 8}

⇒ A ∩ B = {3} ⊂ (C ∩  B) but A ⊂ C is not true.

Hence, statement 2 is false.

Statement 3: (A ∪ B) ⊂ (C ∪ B) for all sets B ⇒ A ⊂ C

This is true as (A ∪ B) ⊂ (C ∪ B) for all sets B ⇒ A ⊂ C

Concept:

If A = [a_ij] is a square matrix of order n and C = [c_ij] is cofactor matrix of A. Then C^T = [c_ji] is called the adjoint of matrix A and is denoted as: adj (A) = C^T = [c_ji], 1 ≤ i, j ≤ n.

Calculation:

Given: \(B = \left[ {\begin{array}{*{20}{c}} 3&2&0\ 2&4&0\ 1&1&0 \end{array}} \right]\)

As we know that, the adj (B) = C^T where C is cofactor matrix of B and it is given by:

\(C = \left[ {\begin{array}{*{20}{c}} {{B_{11}}}&{{B_{12}}}&{{B_{13}}}\ {{B_{21}}}&{{B_{22}}}&{{B_{23}}}\ {{B_{31}}}&{{B_{32}}}&{{B_{33}}} \end{array}} \right];where;{B_{ij}} = {\left( { - 1} \right)^{i + j}} \times {M_{ij}};and;{M_{ij}} = Minor;of;{b_{ij}}\)

⇒ B₁₁ = 0, B₁₂ = 0, B₁₃ = - 2, B₂₁ = 0, B₂₂ = 0, B₂₃ = - 1, B₃₁ = 0, B₃₂ = 0 and B₃₃ = 8.

\(\Rightarrow ;C = \left[ {\begin{array}{{20}{c}} 0&0&{ - 2}\ 0&0&{ - 1}\ 0&0&8 \end{array}} \right];and;adj;\left( B \right) = {C^T} = ;\left[ {\begin{array}{{20}{c}} 0&0&0\ 0&0&0\ { - 2}&{ - 1}&8 \end{array}} \right]\)

Concept:

If \(f\left( x \right) = \left| x \right|,;then;f\left( x \right) = \left{ {\begin{array}{*{20}{c}} { - x,;x < 0}\ {x,;x \ge 0} \end{array}} \right.\)

Calculation:

Given: |x² – x - 6|= x + 2

We can factorize x² – x – 6 as (x - 3) (x + 2)

\(\Rightarrow ;\left| {\left( {x - 3} \right)\left( {x + 2} \right)} \right| = ;\left{ {\begin{array}{*{20}{c}} {\left( {x - 3} \right)\left( {x + 2} \right),;if;\left( {x - 3} \right)\left( {x + 2} \right) \ge 0}\ { - ;\left( {x - 3} \right)\left( {x + 2} \right),;if;\left( {x - 3} \right)\left( {x + 2} \right) < 0} \end{array}} \right.\)

Case – 1: If (x - 3) (x + 2) ≥ 0

⇒ x ∈ (- ∞, - 2] ∪ [3, ∞)

⇒ x² – x – 6 = x + 2

⇒ x² – 2x – 8 = 0

⇒ (x - 4) (x + 2) = 0

⇒ x = 4 or – 2 ∈ (- ∞, - 2] ∪ [3, ∞)

So, the roots of the given quadratic equation are 4 and – 2.

Case – 2: If (x - 3) (x + 2) < 0

⇒ x ∈ [- 2, 3]

⇒ - (x² – x – 6) = x + 2

⇒ x² – 4 = 0

⇒ x = 2 or – 2 ∈ [- 2, 3]

So, the roots of the given quadratic equation are 2 and – 2

Hence, the roots of the given quadratic equation are – 2, 2 and 4.

Concept:

Singular Matrix: Any square matrix of order n is said to be singular if |A| = 0.

Involuntary Matrix: Any square matrix of order n is said to be an involuntary matrix if A² = I, where I is the identity matrix of order n.

Nilpotent Matrix: Any square matrix of order n is said to be nilpotent matrix if there exist least positive integer m such that A^m = O, where O is the null matrix of order n.

Idempotent Matrix: Any square matrix of order n is said to be an idempotent matrix if A² = A.

Calculation:

Given: \(A = \left[ {\begin{array}{*{20}{c}} 0&1\ 1&0 \end{array}} \right],\)

\(\Rightarrow ;{A^2} = ;\left[ {\begin{array}{{20}{c}} 0&1\ 1&0 \end{array}} \right] \times ;\left[ {\begin{array}{{20}{c}} 0&1\ 1&0 \end{array}} \right] \= ;\left[ {\begin{array}{*{20}{c}} 1&0\ 0&1 \end{array}} \right] = I\)

Hence, A is an involuntary matrix as A² = I.

Calculation:

Given: \(\left| {\begin{array}{*{20}{c}} x&-3i&1\ y&1&{i}\ 0&2i&-i \end{array}} \right|=6+11i\)

Let \(A=\left| {\begin{array}{*{20}{c}} x&-3i&1\ y&1&{i}\ 0&2i&-i \end{array}} \right|\)

⇒ det A = x[-i - 2i²] - (-3i)[-yi - 0] + 1[2yi - 0]

As we know that, i2 = -1 

det A = x[-i + 2] + 3i[-yi] + [2yi]

det A = -ix + 2x + 3y + 2yi

det A = (2x + 3y) + (2y - x)i

According to the question, 

det A =  6 + 11i

(2x + 3y) + (2y - x)i = 6 + 11i

⇒ 2x + 3y = 6 and 2y – x = 11

Solving the above equations we get

⇒ x = - 3, y = 4

Concept:

1, ω and ω² are called cube roots of unity where $\omega = \frac{{ - ;1 + i\sqrt 3 }}{2};and;{\omega ^2} = \frac{{ - ;1 - i\sqrt 3 }}{2}$

Properties of the cube roots of unity:

1 + ω + ω² = 0 and ω³ = 1.

Calculation:

Given: The equations z³ + 2z² + 2z + 1 = 0 and z²⁰¹⁷ + z²⁰¹⁸ + 1 = 0.

⇒ z³ + 2z² + 2z + 1 = 0

⇒ (z + 1) × (z² + z + 1) = 0

⇒ z = - 1, ω and ω².

Hence, the roots of the equation z³ + 2z² + 2z + 1 = 0 are – 1, ω and ω².

It is given that the equations z³ + 2z² + 2z + 1 = 0 and z²⁰¹⁷ + z²⁰¹⁸ + 1 = 0 has common roots.

Let f(z) = z²⁰¹⁷ + z²⁰¹⁸ + 1 = z²⁰¹⁷(z + 1) + 1

If z = -1, then f(- 1) = 1 ≠ 0.

Hence, z = - 1 is not a root of f(z).

If z = ω, then f(ω) = ω²⁰¹⁷ (ω + 1) + 1

⇒ f(ω) = (ω³)⁶⁷² × ω × (- ω²) + 1 = 0     ...(∵ 1 + ω + ω² = 0 and ω³ = 1)

Hence, z = ω is a root of f(z).

If z = ω², then f(ω²) = (ω²)²⁰¹⁷ × (ω² + 1) + 1

⇒ f(ω²) = ((ω³)⁶⁷² × ω)² × (- ω) + 1 = 0     ...(∵ 1 + ω + ω² = 0 and ω³ = 1)

Hence, z = ω² is a root of f(z).

So, the common roots of the equations z³ + 2z² + 2z + 1 = 0 and z²⁰¹⁷ + x²⁰¹⁸ + 1 = 0 are ω and ω²

Concept:

The notation C(n, r) is the number of combinations/groups of n different things taking r at a time and is given by: $C;\left( {n,;r} \right) = \frac{{{\rm{n}}!}}{{{\rm{r}}!\left( {{\rm{n;}} - {\rm{;r}}} \right)!}}$

If C(n, x) = C(n, y), then x + y = n.

Calculation:

Given C(20, n + 2) = C(20, n – 2)

As we know that, if C(n, x) = C(n, y), then x + y = n.

⇒ (n + 2) + (n – 2) = 20

⇒ 2n = 20

⇒ n = 10

Concept

If there are n points in a plane and no three points lie on a straight line, then the total number of straight line segments that can be formed is given by: $C\left( {n,;2} \right) = \frac{{{\rm{n}}!}}{{2!\left( {{\rm{n;}} - {\rm{;}}2} \right)!}}$

Calculation

Given: n = 10.

$\Rightarrow ;C\left( {10,;2} \right) = \frac{{10!}}{{2!\left( {10 - {\rm{;}}2} \right)!}} = \frac{{10 \times 9}}{2} = 45$

Hence, the number of straight lines that can be formed is 45.

Concept:

For any quadratic equation, ax² + bx + c = 0. We have discriminant, D = b² – 4ac then the given quadratic equation has:

I. Distinct and real roots if D > 0.

II. Real and repeated roots, if D = 0.

III. Complex roots and conjugate of each other, if D < 0.

Calculation:

In the given quadratic equation px² + qx + r = 0, coefficients p, q, r are positive with roots a and b.

⇒ a + b = - q/p and a × b = r/p.

From the condition: a × b = r/p, we can say that there are two possibilities which are:

Case 1: Both roots are negative so that the product is positive i.e a < 0 and b < 0.

Case 2: Both roots are positive so that the product is positive i.e a > 0 and b > 0.

From the relation a + b = - q/p, we can say that the sum of roots is negative.

Out of the two cases, Case 1 is a valid case satisfying both relations a + b = - q/p and a × b = r/p.

Concept:

Power set: Let A be a non-empty set. Then, \(P\left( A \right) = \left{ {B;|;B; \subseteq A} \right}\) is called as the power set of A.

For any non-empty set A with n(A) = x. Then the total no. of subsets of set A is given by: 2^x.

i.e n(P(A)) = 2^x.

Calculation:

Given: A = {λ, {λ, μ}}

Subsets of A are: {λ}, {{λ, μ}}, {λ, {λ, μ}}, φ

Power set of A, P(A) = {φ, {λ}, {{λ, μ}}, {λ, {λ, μ}}}

Calculation:

In the above Venn diagram, we have represented chess, carom and table tennis with three circles.

Given: 60 play chess, 50 play table tennis, 48 play carom, 12 play chess and carom, 15 play carom and table tennis, 20 play table tennis and chess.

⇒ a + b + d + e = 60     ...1)

⇒ d + e + f + g = 50     ...2)

⇒ b + c + e + f = 48     ...3)

⇒ b + e = 12     ...4)             

⇒ e + f = 15     ...5)

⇒ d + e = 20     ...6)

By using (1) + (2) + (3) – (4) – (5) – (6)

⇒ a + b + c+ d + f + g = 60 + 50 + 48 – 12 – 15 - 20 = 111

We know that, total number of students in school is given by: a + b + c + d + f + g + e = 111 + e

Total number of students in school will be minimum, when e = 0.

⇒ Minimum number of students in school = 111.

Calculation:

⇒ a + b + d + e = 60     ...1)

⇒ d + e + f + g = 50     ...2)

⇒ b + c + e + f = 48     ...3)

⇒ b + e = 12     ...4)

⇒ e + f = 15     ...5)

⇒ d + e = 20     ...6)

By using (1) + (2) + (3) – (4) – (5) – (6)

⇒ a + b + c + d + f + g = 60 + 50 + 48 – 12 – 15 - 20 = 111

Total number of students in school = a + b + c + d + f + g + e = 111 + e.

Maximum number of students will depend on the maximum value of e.

From the equations (4), (5) and (6) we have the following three cases:

Case 1: Let b = 0

⇒ e = 12.

By substituting the value of e in (5), we get

⇒ f = 15 – e = 3

By substituting the value of e in (5), we get

⇒ d = 20 – e = 8

Case 2: Let f = 0

⇒ e = 15.

By substituting the value of e in (4), we get

⇒ b = 12 – e = -3

We got b as negative, so we need to discard this case ∵ b ≥ 0.

Case 3: Let d = 0

⇒ e = 20.

By substituting the value of e in (4), we get

⇒ b = 12 – e = -8

We got b as negative, so we need to discard this case ∵ b ≥ 0.

So the maximum possible value of e is from case 1 i.e. e = 12.

∴ Maximum possible number of students in school = 111 + e = 111 + 12 = 123.

Concept

If A is any matrix of order n and it’s inverse exists, then we can write

AA^-1 = A^-1A = I, where I = Identity matrix of order n

Calculation

Given: A is an identity matrix of order 3 i.e. A = I

Multiplying both sides by A^-1 we get

⇒ AA^-1 = IA^-1

⇒ I = A^-1 [∵ A matrix multiplied by the identity matrix is the matrix itself i.e. AI = A]

⇒ A = A^-1

Concept:

For any square matrix A of order n, |A| = |A^T|

Calculation:

Given: A is a square matrix of order 3 such that its determinate is 4

⇒ |A| = 4

As we know that, for any square matrix A of order n, |A| = |A^T|

⇒ |A^T| = 4

Concept:

The notation C(n, r) is the number of combinations/groups of n different things taking r at a time and is given by: $C\left( {n,;r} \right) = \frac{{{\rm{n}}!}}{{{\rm{r}}!\left( {{\rm{n;}} - {\rm{;r}}} \right)!}}$

Calculation:

Given: 6 programmers and 4 typists, an office wants to recruit 5 people.

Here, n = 10 and r = 5.

Case 1: Group of 5 with 1 typist

1 typist can be chosen from 4 in C(4, 1) = 4 ways

4 programmers can be chosen from 6 in C(6, 4) = 15 ways

⇒ No. of ways to form a group of 5 with 1 typist = 4 × 15 = 60.

Case 2: Group of 5 with 2 typists

2 typist can be chosen from 4 in C(4, 2) = 6 ways

3 programmers can be chosen from 6 in C(6, 3) = 20 ways

⇒ No. of ways to form a group of 5 with 2 typist = 6 × 20 = 120.

Case 3: Group of 5 with 3 typists

3 typist can be chosen from 4 in C(4, 3) = 4 ways

2 programmers can be chosen from 6 in C(6, 2) = 15 ways

⇒ No. of ways to form a group of 5 with 2 typist = 4 × 15 = 60.

Case 4: Group of 5 with 4 typists

4 typist can be chosen from 4 in C(4, 4) = 1 way

1 programmers can be chosen from 6 in C(6, 1) = 6 ways

⇒ No. of ways to form a group of 5 with 2 typist = 1 × 6 = 6.

Total number of ways in forming a group of 5 with at least one typist is = 60 + 120 + 60 + 6 = 246.

Concept:

In the binomial expansion of (a + b)ⁿ there are n + 1 terms.

Calculation:

⇒ {(2x – 3y)² (2x + 3y)²}² = {(2x -3y)× (2x + 3y) × (2x – 3y) × (2x + 3y)}²

As we know that, (a - b) × (a + b) = a² – b²

⇒ {(2x – 3y)² (2x + 3y)²}² = {(4x² – 9y²)²}² = (4x² – 9y²)⁴

As we know that, in the binomial expansion of (a + b)ⁿ there are n + 1 terms.

Here, a = 4x², b = - 9y² and n = 4.

Hence, there are 5 terms in the expansion of {(2x – 3y)² (2x + 3y)²}².

Concept:

The general term in the binomial expansion of (a + b)ⁿ is given by: ${T_{r + 1}} = {;^n}{C_r} \times {a^{n - r}} \times {b^r}$

Calculation:

Given: In the expansion of (1+ ax)ⁿ, the first three terms are respectively 1, 12x and 64x².

2^st term in expansion of (1+ ax)ⁿ: $^n{C_1} \times {\left( 1 \right)^{n - 1}} \times {\left( {ax} \right)^1} = n \times a \times x = 12;x$

⇒ n × a = 12 ⇒ a = 12/n

3^rd term in expansion of (1+ ax)ⁿ: $^n{C_2} \times {\left( 1 \right)^{n - 2}} \times {\left( {ax} \right)^2} = \frac{{n \times \left( {n - 1} \right)}}{2} \times {a^2} \times {x^2} = 64;{x^2}$

$\Rightarrow ;\frac{{n \times \left( {n - 1} \right)}}{2} \times {a^2} = 64$     ...1)

By substituting the value of a in equation (1), we get

$\Rightarrow ;\frac{{n \times \left( {n - 1} \right)}}{2} \times \frac{{144}}{{{n^2}}} = 64$

⇒ 8n = 72 ⇒ n = 9.

Concept:

I. nth term of a GP is given by: T_n = ar^n-1,

Where a = first term and r = common ratio.

II. nth term of an AP is given by: T_n = a + (n - 1) × d,

Where a = first term and d = common difference.

Calculation:

Let us suppose that the numbers 1, 5 and 25 be the p^th, q^th and r^th term respectively of an AP with a = first term and d = common difference.

⇒ a_p = a + (p - 1) × d = 1    ----(1)

⇒ a_q = a + (q - 1) × d = 5    ----(2)

⇒ a_r = a + (r - 1) × d = 25    ----(3)

By subtracting (1) from (2), we get

⇒ (q - p) × d = 4    ----(4)

Similarly, by subtracting (2) from (3), we get

⇒ (r - q) × d = 20    ----(5)

By, dividing (4) and (5), we get

⇒ (r - p)/(q - p) = 5

∵ (r - p)/(q - p) is a rational number and 5 is also a rational number.

⇒ 1, 5 and 25 can be 3 terms of infinite number of AP’s.

So, The option 1st and 2nd are wrong 

Now, Let 1, 5 and 25 are the three terms of a G.P

So, Let P^th, Q^th and R^th term be 1, 5 and 25 resp.

So, a r^P-1 = 1    ----(6)

 a rQ-1 = 5         ----(7) 

a r^R-1 = 25        ----(8)

On doing (7) ÷ (6), we get 

rQ-P = 5        ----(9)

And on doing (8) ÷ (7), we get 

rR-Q = 5       ----(10)

Now, From (9) and(10), we get 

Q - P = R - Q

Q = (P + R)/2

So, there are several or infinite value of P, Q, R which can satisfy the above relation. 

So, There can be infinite number of G.P's.

∴ The numbers 1, 5 and 25 can be three terms of infinite terms of G.P.

Concept:

The nth term of an AP is given by: T_n = a + (n - 1) × d, where a = first term and d = common difference.

Calculation:

As we know that, the nth term of an AP is given by: T_n = a + (n - 1) × d, where a = first term and d = common difference.

Let a be the first term and d is the common difference.

$\Rightarrow ;{a_{p + q}} = a + \left( {p + q - 1} \right) \times d$     ...1)

$\Rightarrow ;{a_{p - q}} = a + \left( {p - q - 1} \right) \times d$     ...2)

By adding (1) and (2), we get

$\Rightarrow ;{a_{p + q}} + {a_{p - q}} = 2;a + 2;\left( {p - 1} \right)d = 2 \times \left[ {a + \left( {p - 1} \right)d} \right] = 2 \times {a_p}$

Concept:

Properties of Determinants

1. A × adj (A) = adj (A) × A = |A| × I, where A is a square matrix of order n and I is the identity matrix of order n.
2. adj (adj.......adj (A))| = ${\left| A \right|^{{{\left( {m - 1} \right)}^n}}}$, where A is a square matrix of order m and n represents the multiplicity of adj.
3. |A × B| = |A| × |B|, where A and B are square matrices of order n.
4. |A| = |A'|, where A' is transpose of matrix A of order n.
5. |- A| = (-1)n × |A|, where A is a square matrix of order n.

Concept:

Minimum value of $m{\sec ^2}\theta + n;cose{c^2}\theta = ;{\left( {\sqrt m + \sqrt n } \right)^2}$

Calculation:

Given: 25 cosec² x + 36 sec² x

As we know that, minimum value of $m{\sec ^2}\theta + n;cose{c^2}\theta = ;{\left( {\sqrt m + \sqrt n } \right)^2}$

Here, m = 36 and n = 25.

∴ The minimum value of 25 cosec² x + 36 sec² x : ${\left( {\sqrt {36} + \sqrt {25} } \right)^2} = 121$

Concept:

If A and B are square matrices of order n, then det (m × AB) = mn × det (A) × det (B) where m ∈ R is a scalar.

Calculation:

Given: A and B be (3 × 3) matrices with det A = 4 and det B = 3

As we know that, if A and B are square matrices of order n, then det (m × AB) = mn × det (A) × det (B) where m ∈ R is a scalar.

⇒ det (2AB) = 23 × det (A) × det (B) = 8 × 4 × 3 = 96.

Concept:

If A and B are square matrices of order n, then det (m × AB) = mn × det (A) × det (B) where m ∈ R is a scalar.

If A is a non – singular square matrix of order n, then det (A-1­) = 1/det (A)

Calculation:

Given: Given: A and B be (3 × 3) matrices with det A = 4 and det B = 3

As we know that, if A and B are square matrices of order n, then det (m × AB) = mn × det (A) × det (B) where m ∈ R is a scalar.

⇒ det (3AB-1) = 33 × det (A) × det (B- 1)

As we know that, If A is a non – singular square matrix of order n, then det (A-1­) = 1/det (A)

$\Rightarrow {\rm{;det}}\left( {3A{B^{ - 1}}} \right) = 27 \times 4 \times \frac{1}{3} = 36$

Concept:

For any complex number, Z = a + ib, modulus of z is given by $r = \sqrt {{a^2} + {b^2}}$

Calculation:

Given: $z = \frac{{1 + 2i}}{{1 - {{\left( {1 - i} \right)}^2}}}$

$\Rightarrow ;z = \frac{{1 + 2i}}{{1 - {{\left( {1 - i} \right)}^2}}} = \frac{{1 + 2i}}{{1 + 2i}} = 1$

Now, we can write z = 1 = 1 + i × 0

On comparing z = 1 + i × 0 with z = a + ib, we get 

⇒ a = 1 and b = 0

As we know that $r = \sqrt {{a^2} + {b^2}}$, where r = |z|

Here, a = 1 and b = 0.

$\therefore ;r = \sqrt {{1^2} + {0^2}} = 1$

Concept:

For any complex number, Z = a + ib, argument of Z is given by: $\arg z = \theta = {\tan ^{ - 1}}\left( {\frac{b}{a}} \right)$ and modulus $r = \sqrt {{a^2} + {b^2}}$

The principal argument of Z is given by: Arg (Z)

Case-1: If the point Z, lies in the first quadrant then, Arg(Z) = θ.

Case-2: If the point Z, lies in the second quadrant then, Arg(Z) = π – θ

Case-3: If the point Z, lies in the third quadrant then, Arg(Z) = θ - π

Case-4: If the point Z, lies in the fourth quadrant then, Arg(Z) = - θ

Note:- Principal value of any complex number Z is unique and - π ≤ Arg (Z) < π.

Calculation:

Given: $z = \frac{{1 + 2i}}{{1 - {{\left( {1 - i} \right)}^2}}}$

$⇒ ;z = \frac{{1 + 2i}}{{1 - {{\left( {1 - i} \right)}^2}}} = \frac{{1 + 2i}}{{1 + 2i}} = 1$

We can write z = 1 as z = 1 + i × 0

On comparing z = 1 + i × 0 with z = a + ib, we get 

⇒ a = 1 and b = 0

As we know that, $\arg z = \theta = {\tan ^{ - 1}}\left( {\frac{b}{a}} \right)$

$⇒ ;\arg z = \theta = {\tan ^{ - 1}}\left( {\frac{0}{1}} \right) = 0$

⇒ z lies on the real axis of the argand plane.

∴  Arg (z) = 0.

Concept:

sin (A + B) = sin A × cos B + sin B × cos A

cos (A - B) = cos A × cos B + sin A × sin B

Calculation:

cos 236° = cos (180° + 56°) = - cos 56°

sin 124° = sin (90° + 34°) = cos 34°

cos 178° = cos (90° + 88°) =- sin 88°

sin 208° = sin (180° + 28°) = - sin 28°

By substituting these values in $\frac{{\sin 34^\circ \cos 236^\circ - \sin 56^\circ \sin 124^\circ }}{{\cos 28^\circ \cos 88^\circ + \cos 178^\circ \sin 208^\circ }}$ we get

$\Rightarrow ;\frac{{\sin 34^\circ \cos 236^\circ - \sin 56^\circ \sin 124^\circ }}{{\cos 28^\circ \cos 88^\circ + \cos 178^\circ \sin 208^\circ }} = \frac{{ - \sin 34^\circ \cos 56^\circ - \sin 56^\circ \cos 34^\circ ;}}{{\cos 28^\circ \cos 88^\circ + ;\sin 88^\circ \sin 28^\circ }}$

As we know that, sin (A + B) = sin A × cos B + sin B × cos A and cos (A - B) = cos A × cos B + sin A × sin B.

$\Rightarrow ;\frac{{\sin 34^\circ \cos 236^\circ - \sin 56^\circ \sin 124^\circ }}{{\cos 28^\circ \cos 88^\circ + \cos 178^\circ \sin 208^\circ }} = - \frac{{\sin \left( {34^\circ + 56^\circ } \right)}}{{\cos \left( {88^\circ - 28^\circ } \right)}} = - 2$

Concept:

$\tan \left( {A + B} \right) = \frac{{\tan A; + \tan B}}{{1 - \tan A\tan B}}$

Calculation:

tan 54° = tan (45° + 9°)

As we know that, $\tan \left( {A + B} \right) = \frac{{\tan A; + \tan B}}{{1 - \tan A\tan B}}$

$\Rightarrow {\rm{;tan}}54^\circ = \tan \left( {45^\circ + 9^\circ } \right) = \frac{{\tan 45^\circ + \tan 9^\circ }}{{1 - \tan 45^\circ \tan 9^\circ }} = \frac{{1 + \tan 9^\circ }}{{1 - \tan 9^\circ }}$

$\Rightarrow {\rm{;tan}}54^\circ = \frac{{\cos 9^\circ + \sin 9^\circ }}{{\cos 9^\circ - \sin 9^\circ }}$

Concept**:**

sin 2A = 2 sin A × cos A

cos 2A = cos² A – sin2 A

Calculation**:**

Given: p = X cos θ - Y sin θ and q = X sin θ + Y cos θ.

⇒ p2 = X² cos² θ + Y² sin² θ – 2 XY cos θ sin θ     ...1)

⇒ q² = X² sin² θ + Y² cos² θ + 2 XY cos θ sin θ     ...2)

⇒ 4pq = 4 [X² cos θ sin θ + XY cos² θ – XY sin² θ – Y² sin θ cos θ]     ...3)

By adding (1), (2) and (3), we get

⇒ p² + 4pq + q² = X² [1 + 2 sin 2θ] + Y² [1 - 2 sin 2θ] + 4 XY cos 2θ

As it is given that, p² + 4pq + q² = AX² + BY²

⇒ X² [1 + 2 sin 2θ] + Y² [1 - 2 sin 2θ] + 4 XY cos 2θ = AX² + BY²

⇒ A = 1 + 2 sin 2θ, B = 1 - 2 sin 2θ and cos 2θ = 0.

⇒ θ = π/4, A = 3 and B = - 1.

Concept**:**

sin 2A = 2 sin A × cos A

cos 2A = cos2 A – sin2 A

Calculation**:**

Given: p = X cos θ - Y sin θ and q = x sin θ + Y cos θ.

⇒ p2 = x2 cos2 θ + Y2 sin2 θ – 2 XY cos θ sin θ     ...1)

⇒ q2 = x2 cos2 θ + Y2 sin2 θ + 2 XY cos θ sin θ     ...2)

⇒ 4pq = 4 [X2 cos θ sin θ + XY cos2 θ – XY sin2 θ – Y2 sin θ cos θ]     ...3)

By adding (1), (2) and (3), we get

⇒ p2 + 4pq + q2 = X2 [1 + 2 sin 2θ] + Y2 [1 - 2 sin 2θ] + 4 XY cos 2θ

As it is given that, p2 + 4pq + q2 = AX2 + BY2

⇒ X2 [1 + 2 sin 2θ] + Y2 [1 - 2 sin 2θ] + 4 XY cos 2θ = AX2 + BY2

⇒ A = 1 + 2 sin 2θ, B = 1 - 2 sin 2θ and cos 2θ = 0.

⇒ θ = π/4, A = 3 and B = - 1.

Concept**:**

sin 2A = 2 sin A × cos A

cos 2A = cos2 A – sin2 A

Calculation**:**

Given: p = X cos θ - Y sin θ and q = x sin θ + Y cos θ.

⇒ p2 = x2 cos2 θ + Y2 sin2 θ – 2 XY cos θ sin θ     ...1)

⇒ q2 = x2 cos2 θ + Y2 sin2 θ + 2 XY cos θ sin θ     ...2)

⇒ 4pq = 4 [X2 cos θ sin θ + XY cos2 θ – XY sin2 θ – Y2 sin θ cos θ]     ...3)

By adding (1), (2) and (3), we get

⇒ p2 + 4pq + q2 = X2 [1 + 2 sin 2θ] + Y2 [1 - 2 sin 2θ] + 4 XY cos 2θ

As it is given that, p2 + 4pq + q2 = AX2 + BY2

⇒ X2 [1 + 2 sin 2θ] + Y2 [1 - 2 sin 2θ] + 4 XY cos 2θ = AX2 + BY2

⇒ A = 1 + 2 sin 2θ, B = 1 - 2 sin 2θ and cos 2θ = 0.

⇒ θ = π/4, A = 3 and B = - 1.

Concept:

sin2 x + cos2 x = 1

cos (A - B) = cos A × cos B + sin A × sin B

Calculation:

Given: cos (θ – α) = a and cos (θ – β) = b.

As we know that, sin2 x + cos2 x = 1

$\Rightarrow ;\sin \left( {\theta - \alpha } \right) = \sqrt {1 - {a^2}} ;and\sin \left( {\theta - \beta } \right) = \sqrt {1 - {b^2}}$

cos (α – β) = cos [(θ - β) – (θ - α)]

As we know that, cos (A - B) = cos A × cos B + sin A × sin B

⇒ cos (α – β) = cos (θ - β) × cos (θ - α) + sin (θ - β) × sin (θ - α)

$\Rightarrow ;\cos ;\left( {\alpha - \beta } \right) = ;ab + \sqrt {1 - {b^2}} \times \sqrt {1 - {a^2}}$

Concept:

sin2 x + cos2 x = 1

cos (A - B) = cos A × cos B + sin A × sin B

sin (A - B) = sin A × cos B – cos A × sin B

Calculation:

Given: cos (θ – α) = a and cos (θ – β) = b.

As we know that, sin2 x + cos2 x = 1

$\Rightarrow ;\sin \left( {\theta - \alpha } \right) = \sqrt {1 - {a^2}} ;and;\sin \left( {\theta - \beta } \right) = \sqrt {1 - {b^2}}$

From previous question, we have $\cos ;\left( {\alpha - \beta } \right) = ;ab + \sqrt {1 - {b^2}} \times \sqrt {1 - {a^2}}$

sin (α – β) = sin [(θ - β) – (θ - α)]

As we know that, sin (A - B) = sin A × cos B – cos A × sin B

⇒ sin (α – β) = sin (θ - β) × cos (θ - α) - cos (θ - β) × sin (θ - α)

$\Rightarrow ;\sin ;\left( {\alpha - \beta } \right) = a\sqrt {1 - {b^2}} - b\sqrt {1 - {a^2}}$

$\Rightarrow ;{\sin ^2}\left( {\alpha - \beta } \right) + 2;ab\cos \left( {\alpha - \beta } \right) = {a^2} + {b^2}$

Concept:

sin² x + cos² x = 1

sin 2x = 2 sin x cos x

Calculation:

Given: sin α + cos α = p

By squaring both the sides, we get

⇒ sin² α + cos² α + 2 sin α cos α = p²

As we know that, sin² x + cos² x = 1 and sin 2x = 2 sin x cos x

⇒ 1 + sin 2α = p²

⇒ sin 2α = p² – 1

As we can write cos² 2α = 1 – sin² 2α

⇒ cos2 2α = 1 – (p² – 1)²

∴ The value of cos² 2α is p² (2 – p²).

Concept:

sin^-1 (x) = cosec^-1 (1/x)

sec^-1 y + cosec^-1 y = π/2, y ≤ - 1 or y ≥ 1.

Calculation:

Given: ${\sin ^{ - 1}}\frac{4}{5} + {\sec ^{ - 1}}\frac{5}{4} - \frac{\pi }{2}$

As we know that, sin^-1 (x) = cosec^-1 (1/x)

$\Rightarrow ;{\sin ^{ - 1}}\frac{4}{5} + {\sec ^{ - 1}}\frac{5}{4} - \frac{\pi }{2} = cose{c^{ - 1}};\left( {\frac{5}{4}} \right) + ;{\sec ^{ - 1}}\frac{5}{4} - \frac{\pi }{2}$

As we know that, sec^-1 y + cosec^-1 y = π/2, y ≤ - 1 or y ≥ 1

$\Rightarrow ;{\sin ^{ - 1}}\frac{4}{5} + {\sec ^{ - 1}}\frac{5}{4} - \frac{\pi }{2} = \frac{\pi }{2} - \frac{\pi }{2} = 0$

Concept:

$\sin 2\theta = \frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}$

$\cos 2\theta = \frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$

$\tan 2\theta = \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$

${\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x; - ;y}}{{1 + xy}}} \right),;if;x > 0,;y > 0$

Calculation:

Given: ${\sin ^{ - 1}}\frac{{2p}}{{1 + p2}} - {\cos ^{ - 1}}\frac{{1 - {q^2}}}{{1 + {q^2}}} = {\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}}$

Let, p = tan A, q = tan B and x = tan C

$\Rightarrow {\rm{;si}}{{\rm{n}}^{ - 1}}\frac{{2\tan A}}{{1 + {{\tan }^2}A}} - {\cos ^{ - 1}}\frac{{1 - {{\tan }^2}B}}{{1 + {{\tan }^2}B}} = {\tan ^{ - 1}}\frac{{2\tan C}}{{1 - {{\tan }^2}C}}$

$\Rightarrow ;2A - 2B = 2C$

⇒ A – B = C

⇒ tan^-1 p – tan^-1 q = tan^-1 x

As we know that, ${\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x; - ;y}}{{1 + xy}}} \right)$

$\Rightarrow ;{\tan ^{ - 1}}p - {\tan ^{ - 1}}q = {\tan ^{ - 1}}\left( {\frac{{p - q}}{{1 + pq}}} \right) = {\tan ^{ - 1}}x$

$\Rightarrow ;x = \frac{{p - q}}{{1 + pq}}$

Concept:

$\tan \left( {A + B} \right) = \frac{{\tan A + \tan B}}{{1 - \tan A \times \tan B}}$

Calculation:

Given: $\tan \theta = \frac{1}{2}$ and $\tan \varphi = \frac{1}{3}$

As we know that, $\tan \left( {A + B} \right) = \frac{{\tan A + \tan B}}{{1 - \tan A \times \tan B}}$

$\Rightarrow ;\tan \left( {\theta + \phi } \right) = \frac{{\tan \theta + \tan \phi }}{{1 - \tan \theta \tan \phi }} = \frac{{\frac{1}{2} + \frac{1}{3}}}{{1 - \frac{1}{6}}} = 1 = \tan \frac{\pi }{4}$

⇒ θ + φ = π/4

Concept:

sin² x + cos² x = 1

2 sin A sin B = cos (A - B) – cos (A + B)

cos 2A = cos² A – sin² A

Calculation:

Given: $cosA = \frac{3}{4}$

As we know that, sin² x + cos² x = 1

$\Rightarrow ;{\sin ^2}A = 1 - {\cos ^2}A = 1 - \frac{9}{{16}} = \frac{7}{{16}}$

$\Rightarrow ;\sin A = \frac{{\sqrt 7 }}{4}$

$\Rightarrow ;\sin \left( {\frac{A}{2}} \right)\sin ;\left( {\frac{{3A}}{2}} \right) = \frac{1}{2} \times \left[ {2 \times \sin \left( {\frac{A}{2}} \right)\sin ;\left( {\frac{{3A}}{2}} \right)} \right]$

As we know that, 2 sin A sin B = cos (A - B) – cos (A + B)

$\Rightarrow ;\sin \left( {\frac{A}{2}} \right)\sin ;\left( {\frac{{3A}}{2}} \right) = \frac{1}{2} \times (\cos A - \cos 2A)$      ...1)

As we know that, cos 2A = cos² A – sin² A

$\Rightarrow ;\cos 2A = \frac{9}{{16}} - \frac{7}{{16}} = \frac{1}{8}$

By substituting the value of cos A and cos 2A in equation (1), we get

$\Rightarrow ;\sin \left( {\frac{A}{2}} \right)\sin \left( {\frac{{3A}}{2}} \right) = \frac{1}{2} \times \left( {\frac{3}{4} - \frac{1}{8}} \right) = \frac{5}{{16}}$

Concept:

$\tan \left( {A + B} \right) = \frac{{\tan A + \tan B}}{{1 - \tan A \times \tan B}}$

Calculation:

Given: tan 75° + cot 75°

⇒ tan 75° = tan (30° + 45°)

As we know that, $\tan \left( {A + B} \right) = \frac{{\tan A + \tan B}}{{1 - \tan A \times \tan B}}$

$\Rightarrow ;\tan 75^\circ = \tan \left( {45^\circ + 30^\circ } \right) = \frac{{\tan 45^\circ + \tan 30^\circ }}{{1 - \tan 45^\circ \tan 30^\circ }} = \frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}$

As we know that, cot x = 1/tan x

$\Rightarrow ;\cot 75^\circ = \frac{1}{{\tan 75^\circ }} = \frac{{\sqrt 3 ; - ;1}}{{\sqrt 3 ; + ;1}}$

⇒ tan 75° + cot 75° = 4

Calculation:

Given: cos 46° cos 47° cos 48° cos 49° cos 50° ….

As we know that, cos 90° = 0.

⇒ cos 46° cos 47° cos 48° cos 49° cos 50° …. = cos 46° cos 47° cos 48° cos 49° cos 50° … cos 90°… = 0.

Concept:

$\sin 18^\circ = \frac{{\sqrt 5 - 1}}{4}$

If sin x = sin y then x = n π + (- 1)ⁿ y

Calculation:

Given: sin 2θ = cos 3θ, where $0 < \theta < \frac{\pi }{2}$

As we know that, cos x = sin (90° - x)

$\Rightarrow ;\sin 2\theta = \sin \left( {\frac{\pi }{2} - 3\theta } \right)$

As we know that, if sin x = sin y then x = n π + (- 1)ⁿ y

$\Rightarrow ;2\theta = n\pi + {\left( { - 1} \right)^n};\left( {\frac{\pi }{2} - 3\theta } \right)$

$0 < \theta < \frac{\pi }{2}$

$\Rightarrow ~2\theta =\frac{\pi }{2}-~3\theta \Rightarrow ~\theta =\frac{\pi }{10}$

$\Rightarrow ;\sin \left( {\frac{\pi }{{10}}} \right) = \sin 18^\circ = \frac{{\sqrt 5 - 1}}{4}$

Concept:

If α and β are the roots of the quadratic equation ax² + bx + c = 0. Then

$\alpha + \beta = - \frac{b}{a};and;\alpha \times \beta = \frac{c}{a}$

$\tan \left( {A + B} \right) = \frac{{\tan A + \tan B}}{{1 - \tan A \times \tan B}}$

Calculation:

Given: tan 19° and tan 26° are the roots of the equation x² + px + q = 0.

As we know that, if α and β are the roots of the quadratic equation ax² + bx + c = 0. Then

$\alpha + \beta = - \frac{b}{a};and;\alpha \times \beta = \frac{c}{a}$

Here, a = 1, b = p and c = q.

⇒ tan 19° + tan 26° = - p and tan 19° × tan 26° = q.

As we know that, $\tan \left( {A + B} \right) = \frac{{\tan A + \tan B}}{{1 - \tan A \times \tan B}}$

$\Rightarrow ;\tan \left( {19^\circ + 26^\circ } \right) = \frac{{\tan 19^\circ + \tan 26^\circ }}{{1 - \tan 19^\circ \times \tan 26^\circ }} = \tan 45^\circ = 1$

$\Rightarrow ;\frac{{ - ;p}}{{1 - q}} = 1 \Rightarrow q - p = 1$

Concept:

If a_n is the nth term of an AP and S_n is the sum of n terms of the same AP then a_n = S_n – S_{n - 1}

Calculation:

Given: S_n = n (n + 1)

⇒ S_{n - 1} = (n - 1) × (n) = n² - n

⇒ S_n = n² + n

As we know that, if a_n is the nth term of an AP and S_n is the sum of n terms of the same AP then a_n = S_n – S_{n - 1}

⇒ a_n = S_n – S_{n – 1} = 2n

⇒ a₄ = 2 × 4 = 8

Concept:

1 + tan² x = sec² x

(a + b)² = a² + b² + 2ab

Calculation:

⇒ (1 + tan α tan β)² = 1 + tan² α tan² β + 2 tan α tan β     ...1)

⇒ (tan α – tan β)² = tan² α + tan² β – 2 tan α tan β           ...2)

Adding (1) and (2), we get

⇒ (1 + tan α tan β)² + (tan α – tan β)²

⇒ 1 + tan2 α tan2 β + 2 tan α tan β + tan2 α + tan2 β – 2 tan α tan β 

⇒ 1 + tan2 α tan2 β + tan2 α + tan2 β

⇒ 1 + tan2 α + tan2 α tan2 β + tan2 β

⇒ (1 + tan2 α) + tan2 β (1 + tan2 α)

⇒ (1 + tan2 α)(1 + tan2 β)

⇒ sec² α sec² β

∴ (1 + tan α tan β)² + (tan α – tan β)² – sec² α sec² β is equal to 0.