Quantitative Aptitude PYQs

CONCEPT:

• i² = -1
• i⁴ⁿ = 1

 

CALCULATION:

Given:

By rationalising with (1+i)

\(\Rightarrow {\left( {\frac{{\left( {1 + i} \right)\left( {1 + i} \right)}}{{\left( {1 - i} \right)\left( {1 + i} \right)}}} \right)^n} = {\left( {\frac{{1 + 2i + {i^2}}}{{1 - {i^2}}}} \right)^n}\) {∵ (A+ B) (A – B) = A² – B²)}

\(\Rightarrow {\left( {\frac{{1 + 2i - 1}}{{1 + 1}}} \right)^n} = {\left( {\frac{{2i}}{2}} \right)^n}\) {∵ i² = -1}

⇒ iⁿ = 1

The smallest value of ‘n’ for which iⁿ = 1 is 4.

CONCEPT:

Equivalence Relation:

Any relation R on a non-empty set A which is reflexive, symmetric and transitive is called an equivalence relation on A.

Reflexive Relation:

For all a ∈ A, if (a, a) ∈ R, then R is called a reflexive relation on A.

Symmetric Relation:

For all a, b ∈ A, if (a, b) ∈ R⇒ (b, a) ∈ R then R is called a symmetric relation on A.

Transitive Relation:

For any a, b, c ∈ A if (a, b) ∈ R, (b, c) ∈ R ⇒ (a, c) ∈ R, then R is called a transitive relation on A.

 

CALCULATION:

Given that: (a, b) R (c, d) ⇔ a + d = b + c

Let (a, b) ∈ N × N

⇒ (a + b) = (b + a)

⇒ (a, b) R (a, b)

Hence, R is reflexive.

Let (a, b) R (c, d), where (a, b), (c, d) ∈ N × N

⇒ a + d = b + c

⇒ b + c = a + d

⇒ c + b = a + d

⇒ (c, d) R (a, b)

Hence, R is symmetric.

Given that: (a, b) R (c, d)

⇒ a + d = b + c

⇒ a – b = c – d … equation (1)

Let us suppose (c, d) R (e, f)

⇒ c + f = d + e

⇒ c – d = e – f …equation (2)

From equations (1) and (2)

⇒ a – b = e – f

⇒ a + f = b + e

⇒ (a, b) R (e, f)

Hence, R is transitive.

Relation R satisfies reflexive, symmetric and transitive.

Hence, R is an equivalence relation.

CONCEPT:

Infinite Geometric series:

Standard infinite geometric series is given by: a , ar , ar² , ar³ ,……,ar^n-1 , ….up to infinite terms

Sum of infinite terms of Geometric Progression is given by: \({{\rm{s}}_{\rm{n}}} = \frac{{\rm{a}}}{{1 - {\rm{r}}}}\)

CALCULATION:

Given that: y = x + x² + x³ + …… up to infinite terms

Comparing it with standard infinite G.P series, we get a = x and r = x

Sum of infinite terms of geometric progression is given by: \({{\rm{s}}_{\rm{n}}} = \frac{{\rm{a}}}{{1 - {\rm{r}}}}\)

\(\Rightarrow \;{{\rm{s}}_{\rm{n}}} = \left( {\frac{{\rm{x}}}{{1 - {\rm{x}}}}} \right)\)

\(\Rightarrow {\rm{y}} = {\rm{}}\left( {\frac{{\rm{x}}}{{1 - {\rm{x}}}}} \right)\;\)

⇒ y (1 – x) = x

⇒ y – xy = x

⇒ y = x + xy

⇒ y = x(1+y)

\(\Rightarrow x = \left( {\frac{y}{{1 + y}}} \right)\)

CONCEPT:

If α and β are the roots of ax² + bx + c = 0, then the equation can be written as: a (x² – (α +β) x + α β) = 0 i.e., (x - α) (x - β) = 0 

CALCULATION:

Given that: α and β are the roots of the equation 3x² + 2x + 1 = 0

\(\Rightarrow {\rm{\alpha }} + {\rm{\beta }} = \frac{{ - 2}}{3}\) and \({\rm{\alpha }} \times {\rm{\beta }} = \frac{1}{3}\)

We know that, the standard quadratic equation is given by: x² - (sum of roots) x + (product of roots) = 0

Sum of roots is given by:

\(\Rightarrow \left( {{\rm{\alpha }} + {{\rm{\beta }}^{ - 1}}} \right) + \left( {{\rm{\beta }} + {{\rm{\alpha }}^{ - 1}}} \right) = {\rm{\alpha }} + {\rm{\beta }} + \left( {\frac{1}{{\rm{\alpha }}}} \right) + \left( {\frac{1}{{\rm{\beta }}}} \right) = {\rm{\alpha }} + {\rm{\beta }} + \left( {\frac{{{\rm{\alpha }} + {\rm{\beta }}}}{{{\rm{\alpha \beta }}}}} \right)\)

\(\Rightarrow \left( {{\rm{\alpha }} + {{\rm{\beta }}^{ - 1}}} \right) + \left( {{\rm{\beta }} + {{\rm{\alpha }}^{ - 1}}} \right) = \frac{{ - 2}}{3} + \frac{{\frac{{ - 2}}{3}}}{{\frac{1}{3}}} = \frac{{ - 2}}{3} + \left( { - 2} \right) = \frac{{ - 8}}{3}\)

Product of the roots is given by:

\(\Rightarrow \left( {{\rm{\alpha }} + {{\rm{\beta }}^{ - 1}}} \right)\left( {{\rm{\beta }} + {{\rm{\alpha }}^{ - 1}}} \right) = \left( {{\rm{\alpha \beta }} + 1 + 1 + \frac{1}{{{\rm{\alpha \beta }}}}} \right) = \frac{1}{3} + 1 + 1 + 3 = \frac{{16}}{3}\)

The quadratic equation whose roots are α + β^-1 and β + α^-1 is given by:

\(\Rightarrow {x^2} - \left( {\frac{{ - 8}}{3}} \right)x + \frac{{16}}{3} = 0\)

\(\Rightarrow {x^2} + \frac{8}{3}x + \frac{{16}}{3} = 0\)

⇒ 3x²+ 8x + 16 = 0

CONCEPT:

Formulas and Properties of Logarithms:

1. \({\log _{\rm{a}}}{{\rm{m}}^{\rm{n}}} = {\rm{\;nlo}}{{\rm{g}}_{\rm{a}}}{\rm{m}}\)
2. \(\frac{1}{{{{\log }_a}m}} = \;{\log _m}a\)

 

CALCULATION:

Given that: \(\frac{1}{{{{\log }_3}e}} + \frac{1}{{{{\log }_3}{e^2}}} + \frac{1}{{{{\log }_3}{e^4}}} + \ldots \) up to infinite terms

\(\Rightarrow \frac{1}{{{{\log }_3}{\rm{e}}}} + \frac{1}{{{{\log }_3}{{\rm{e}}^2}}} + \frac{1}{{{{\log }_3}{{\rm{e}}^4}}} + \ldots = \frac{1}{{{{\log }_3}{\rm{e}}}} + \frac{1}{{2{{\log }_3}{\rm{e}}}} + \frac{1}{{4{{\log }_3}{\rm{e}}}} + ......\)

\(\Rightarrow \frac{1}{{{{\log }_3}{\rm{e}}}} + \frac{1}{{{{\log }_3}{{\rm{e}}^2}}} + \frac{1}{{{{\log }_3}{{\rm{e}}^4}}} + \ldots = \frac{1}{{{{\log }_3}{\rm{e}}}}\left[ {1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots } \right]\)

\(\left\{\because {{{\log }_{\rm{a}}}{{\rm{m}}^{\rm{n}}} = {\rm{\;nlo}}{{\rm{g}}_{\rm{a}}}{\rm{m}}} \right\}\)

The series: \(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + {\rm{\;}} \cdots {\rm{\;}}\)is in the form of infinite geometric series.

Comparing it with standard infinite G.P series, we get a = 1 and r = 1 / 2

Sum of infinite terms of Geometric Progression is given by: \({{\rm{s}}_{\rm{n}}} = \frac{{\rm{a}}}{{1 - {\rm{r}}}}\)

\(\Rightarrow {{\rm{s}}_{\rm{n}}} = {\rm{\;}}\frac{1}{{1 - \frac{1}{2}}} = 2\)

As we know that, \(\frac{1}{{{{\log }_a}{\rm{m}}}} = {\rm{\;log}}_{\rm{m}}{\rm{a}}\), by using this property we get

\(\Rightarrow \frac{1}{{{\rm{log}}_3{\rm{e}}}}\left[ {1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots } \right] = \frac{2}{{{\rm{log}}_3{\rm{e}}}} = 2{\rm{log}}_{\rm{e}}3 = \;{\rm{log}}_{\rm{e}}9\)

CONCEPT:

Permutations:

The number of permutations of n distinct things taken r at a time is denoted by ⁿP_r = n! /(n-r)!

CALCULATION:

Let us consider the below diagram as a long table which contain two sides named as “A” side, “B” side.

Given that: 4 particular men wish to sit on one particular side and 2 particular men on the other side.

So, let us suppose that 4 particular men sat on “A” side and the other 2 sat on “B” side.

The number of ways in which 4 people can sit in 8 chairs on “A” side are ⁸P₄ ways and the number of ways in which 2 people can sit in 8 chairs on “B” side are ⁸P₂ ways.

Now, we are left with 10 people and there are 10 chairs remaining.

Hence, the total number of ways of arranging 10 people in 10 chairs are 10! Ways.

Hence, the required answer is ⁸P₄ × ⁸P_2­ × 10!  = 210 × 8! × 8! Ways

Concept

Let the system of equations be,

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃

\(\; \Rightarrow \left[ {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\\ {{a_2}}&{{b_2}}&{{c_2}}\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right]\;\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \;\left[ {\begin{array}{*{20}{c}} {{d_1}}\\ {{d_2}}\\ {{d_3}} \end{array}} \right]\)

⇒ AX = B

⇒ X = A^-1 B = \(\frac{{{\rm{adj\;}}\left( {\rm{A}} \right)}}{{\det {\rm{\;}}({\rm{A}})}}\;B\)

⇒ If det (A) ≠ 0, the system is consistent having unique solution.

⇒ If det (A) = 0 and (adj A). B = 0, system is consistent, with infinitely many solutions.

⇒ If det (A) = 0 and (adj A). B ≠ 0, system is inconsistent (no solution)

Calculation:

Given: 

kx + y + z = 1, x + ky + z = k and x + y + kz = k²

\( \Rightarrow {\rm{A}} = {\rm{}}\left[ {\begin{array}{*{20}{c}} {\rm{k}}&1&1\\ 1&{\rm{k}}&1\\ 1&1&{\rm{k}} \end{array}} \right],{\rm{B}} = {\rm{}}\left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right]{\rm{and\;C}} = \left[ {\begin{array}{*{20}{c}} 1\\ {\rm{k}}\\ {{{\rm{k}}^2}} \end{array}} \right]\)

⇒ For the given equations to have no solution, |A| = 0

\(\Rightarrow \left| {\begin{array}{*{20}{c}} {\rm{k}}&1&1\\ 1&{\rm{k}}&1\\ 1&1&{\rm{k}} \end{array}} \right| = 0\)

⇒ k(k² – 1) - 1(k – 1) + 1(1 – k) = 0

⇒ k³ – k – k + 1 + 1 – k = 0

⇒ k³ -3k +2 = 0

⇒ (k – 1) (k – 1) (k + 2) = 0

⇒ k = 1, -2

If we put k = 1 in the above given equations, then all the equations will become the same.

Hence, the given equations have no solution if k = - 2.

CONCEPT:

Geometric Progression:

The standard form of geometric progression is given by: a, ar, ar², ar³, …., arⁿ

The sum of n terms of a Geometric progression is given by: \({{\rm{s}}_{\rm{n}}} = \frac{{{\rm{a}}\left( {{{\rm{r}}^{\rm{n}}} - 1} \right)}}{{{\rm{r}} - 1}}\) if r≠ 1

CALCULATION:

Given that: \(1.3 + {2.3^2} + \ldots + {\rm{n}}{.3^{\rm{n}}} = \frac{{\left( {\left( {2{\rm{n}} - 1} \right){3^{\rm{a}}} + {\rm{b}}} \right)}}{4}{\rm{\;}}\)

Let L.H.S = \(1.3 + {2.3^2} + \ldots + {\rm{n}}{.3^{\rm{n}}}\) = S

Let us put x = 3

⇒ L.H.S = s = 1.x + 2.x² + ……+ n.xⁿ … equation (1)

Multiply equation (1) with ‘x’

⇒ xs = 1.x² + 2.x³ + …… +n.xⁿ⁺¹….equation(2)

Equation (1) – equation (2) ⇒ s – xs = (1.x + 2.x² + ……+ n.xⁿ) – (1.x² + 2.x³ + …… +n.xⁿ⁺¹)

⇒ s (1 – x) = x + x² + x³ + ……+xⁿ – n.xⁿ⁺¹

\(\Rightarrow {\rm{s}}\left( {1 - {\rm{x}}} \right) = \frac{{{\rm{x}}\left( {1 - {{\rm{x}}^{\rm{n}}}} \right)}}{{1 - {\rm{x}}}} - {\rm{n}}.{{\rm{x}}^{{\rm{n}} + 1}}\)   {∵ \({{\rm{s}}_{\rm{n}}} = \frac{{{\rm{a}}\left( {{{\rm{r}}^{\rm{n}}} - 1} \right)}}{{{\rm{r}} - 1}}\)  }

\(\Rightarrow {\rm{s}} = \frac{1}{{\left( {{\rm{x}} - 1} \right)}}\left[ {\frac{{ - {\rm{x}}\left( {{{\rm{x}}^{\rm{n}}} - 1} \right) + {\rm{n}}.{{\rm{x}}^{{\rm{n}} + 1}}\left( {{\rm{x}} - 1} \right)}}{{{\rm{x}} - 1}}} \right]\)

Put x = 3

\(\Rightarrow {\rm{s}} = \frac{1}{2}\left( {\frac{{ - 3\left( {{3^{\rm{n}}} - 1} \right) + {\rm{n}}{{.3}^{{\rm{n}} + 1}}\left( 2 \right)}}{2}} \right) = {\rm{\;}}\frac{{{3^{{\rm{n}} + 1}}\left( {2{\rm{n}} - 1} \right) + 3}}{4}\)

Comparing L.H.S with R.H.S we get,

a = n+1, b = 3.

CONCEPT:

Trigonometric Identity:

\(\tan \left( {{\rm{\alpha }} \pm {\rm{\beta }}} \right) = {\rm{\;}}\frac{{{\rm{tan\alpha }} \pm {\rm{tan\beta }}}}{{1 \mp {\rm{tan\alpha tan\beta }}}}\)

CALCULATION:

Given that: \(\tan \left( {\frac{P}{2}} \right)\) and \(\tan \left( {\frac{Q}{2}} \right)\) are the roots of the equation ax² + bx + c = 0,\(\angle R = \frac{\pi }{2}\)

\(\Rightarrow \tan \left( {\frac{{\rm{P}}}{2}} \right) + \tan \left( {\frac{{\rm{Q}}}{2}} \right) = \frac{{ - {\rm{b}}}}{{\rm{a}}}\)

\(\Rightarrow \tan \left( {\frac{{\rm{P}}}{2}} \right) \times \tan \left( {\frac{{\rm{Q}}}{2}} \right) = \frac{{\rm{c}}}{{\rm{a}}}\)

We know that, Sum of interior angles of a triangle = π

⇒ ∠ P + ∠ Q + ∠ R = π

⇒ ∠ P + ∠ Q = π /2

We know that, \(\tan \left( {{\rm{\alpha }} \pm {\rm{\beta }}} \right) = {\rm{\;}}\frac{{{\rm{tan\alpha }} \pm {\rm{tan\beta }}}}{{1 \mp {\rm{tan\alpha tan\beta }}}}\)

Now replace α = P/2 and β = Q/2

\(\Rightarrow \tan \left( {\frac{{\rm{P}}}{2} + \frac{{\rm{Q}}}{2}} \right) = \frac{{\tan \left( {\frac{{\rm{P}}}{2}} \right) + \tan \left( {\frac{{\rm{Q}}}{2}} \right)}}{{1 - {\rm{\;}}\tan \left( {\frac{{\rm{P}}}{2}} \right)\tan \left( {\frac{{\rm{Q}}}{2}} \right)}}\)

\(\Rightarrow \tan \left( {\frac{{{\rm{P}} + {\rm{Q}}}}{2}} \right) = \;\frac{{\frac{{ - b}}{a}}}{{1 - \frac{c}{a}}}\;\left\{ \because {\tan \left( {\frac{{\rm{P}}}{2}} \right) + \tan \left( {\frac{{\rm{Q}}}{2}} \right) = \frac{{ - {\rm{b}}}}{{\rm{a}}}\;and{\rm{\;tan}}\left( {\frac{{\rm{P}}}{2}} \right) \times \tan \left( {\frac{{\rm{Q}}}{2}} \right) = \frac{{\rm{c}}}{{\rm{a}}}} \right\}\)

\(\Rightarrow \tan \left( {\frac{\pi }{4}} \right) = \;\frac{{ - b}}{{a - c}}\)  {∵ ∠ P + ∠ Q = π /2}

\(\Rightarrow 1 = \;\frac{{ - b}}{{a - c}}\left\{\because {\tan \left( {\frac{\pi }{4}} \right) = 1} \right\}\)

⇒ a – c = (-b)

⇒ a + b = c

CONCEPT:

Properties of Modulus Function:

1. |ab| = |a||b|
2. |a + b| ≤ |a| + |b|
3. |a - b| ≤ |a| + |b|
4. |a - b| ≥ ||a| - |b||

Quadratic Equation:

The standard quadratic equation is given by: f(x) = ax² + bx + c = 0

The roots of a quadratic equation can be calculated by making f(x) = 0

Thus, the roots of a quadratic function are given by: \({\rm{x}} = \frac{{ - {\rm{b\;}} \pm {\rm{\;}}\sqrt {{{\rm{b}}^2} - 4{\rm{ac}}} }}{{2{\rm{a}}}}{\rm{\;}}\)

CALCULATION:

Given that \(\left| {z - \frac{4}{z}} \right| = 2\)

\(\left| Z \right| = \left| {z - \frac{4}{z} + \frac{4}{z}} \right|\)

Let us consider \(z - \frac{4}{z}\) as ‘a’ and

 as ‘b’

We know that, |a + b| ≤ |a| + |b|

\(\Rightarrow \left| {z - \frac{4}{z} + \frac{4}{z}} \right| \le \left| {z\; - \frac{4}{z}} \right| + \left| {\frac{4}{z}} \right|\)

\( \Rightarrow \left| z \right| \le 2 + \left| {\frac{4}{z}} \right|\)   {∵ \(\left| {z - \frac{4}{z}} \right| = 2\) }

\(\Rightarrow \left| z \right| - \;2 - \left| {\frac{4}{z}} \right| \le 0\)

\(\Rightarrow \frac{{{{\left| z \right|}^2} - 2\left| z \right| - 4}}{{\left| z \right|}} \le 0\)

\(\Rightarrow {\left| {\rm{z}} \right|^2} - 2\left| {\rm{z}} \right| - 4 \le 0\) (∵ |z| > 0)

Finding Roots of |z|² - 2|z| - 4 = 0

⇒ |z| = \(\frac{{ - \left( { - 2} \right)\; \pm \;\sqrt {{{\left( { - 2} \right)}^{2\;}} - \;4\; \times \;1\; \times \; - 4} }}{{2 \times 1}} = \frac{{2\; \pm \;\sqrt {20} }}{2} = 1 \pm \sqrt 5 \)

The roots of the equation \(\frac{{{{\left| z \right|}^2} - 2\left| z \right| - 4}}{{\left| z \right|}} \le 0\) are given by:|Z| = 1± √5

Hence, the maximum value of |Z| is given by: 1 + √5

CONCEPT:

Trigonometric Identities:

• \({\rm{sin\theta }} = \frac{{{\rm{opposite\;side\;}}}}{{{\rm{hypotenuse\;}}}}\)
• \({\rm{cos\theta }} = \frac{{{\rm{adjacent\;side\;}}}}{{{\rm{hypotenuse\;}}}}\)
• \({\rm{tan\theta }} = \frac{{{\rm{opposite\;side\;}}}}{{{\rm{\;adjacent\;side\;}}}}\)
• \(\tan \left( {{\rm{\alpha }} \pm {\rm{\beta }}} \right) = \frac{{{\rm{tan\alpha }} \pm {\rm{tan\beta }}}}{{1 \mp {\rm{tan\alpha tan\beta }}}}\)

 

CALCULATION:

AB = DE = 25 m

AE = PE = x m

As we know that, tan θ = opposite side / adjacent side

In Δ AEC, \(\tan \left( {15^\circ } \right) = \frac{{h - 25}}{x}\)

As we know that,

\(\Rightarrow \tan \left( {15^\circ } \right) = \tan \left( {45^\circ - 30^\circ } \right) = \;\frac{{1 - \frac{1}{{\sqrt 3 }}}}{{1 + \frac{1}{{\sqrt 3 }}}} = \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}\)

\(\Rightarrow \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} = {\rm{\;}}\frac{{{\rm{h}} - 25}}{{\rm{x}}}\)        …equation (1)

In Δ APE, tan (45°) = AP/PE

\(\Rightarrow \frac{{25 + {\rm{h\;}}}}{{\rm{x}}} = 1\)

\(\Rightarrow {\rm{x}} = 25 + {\rm{h\;}}\)       …equation (2)

Substituting equation (2) in equation (1)

\(\Rightarrow \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} = \frac{{{\rm{h}} - 25}}{{{\rm{h}} + 25}}\)

⇒ (√3 – 1) (h + 25) = (√3 +1) (h – 25)

⇒ 50√3 = 2h

⇒ h = 25√3 m

CONCEPT:

Trigonometric Identities:

• Sin (90° - θ) = cos θ
• Cos (90° - θ) = sin θ
• tan (90° - θ) = cot θ
• \({\rm{tan\theta }} = {\rm{\;}}\frac{{{\rm{sin\theta }}}}{{{\rm{cos\theta }}}}\)
• \(\cot {\rm{\theta }} = {\rm{\;}}\frac{{{\rm{cos\theta }}}}{{{\rm{sin\theta }}}}\)
• sin² θ + cos² θ = 1
• 2sinθcosθ = sin2θ
• \({\rm{sin}}18^\circ = {\rm{\;}}\frac{{\sqrt 5 - 1}}{4}\)
• \({\rm{sin}}54^\circ = \frac{{\sqrt 5 + 1}}{4}\)

 

CALCULATION:

Given that, tan 9° - tan 27° - tan 63° + tan 81°

\(\Rightarrow \tan 9^\circ - \tan 27^\circ - \tan \left( {90 - 27} \right)^\circ + \tan \left( {{\rm{\;}}90 - 9} \right)^\circ \)

As we know that,

\(\begin{array}{l} \Rightarrow \tan 9^\circ - \tan 27^\circ - \cot 27^\circ + \cot 9^\circ \\ \Rightarrow \tan 9^\circ + \cot 9^\circ - (\tan 27^\circ + \cot 27^\circ ) \end{array}\)

As we know that, \({\rm{tan\theta }} = {\rm{\;}}\frac{{{\rm{sin\theta }}}}{{{\rm{cos\theta }}}}{\rm{\;and}}\cot {\rm{\theta }} = {\rm{\;}}\frac{{{\rm{cos\theta }}}}{{{\rm{sin\theta }}}}\)

\(\Rightarrow \frac{{\sin 9\;^\circ }}{{\cos 9\;^\circ }} + \frac{{\cos 9^\circ }}{{\sin 9^\circ }} - \left( {\frac{{\sin 27^\circ }}{{\cos 27^\circ }} + \frac{{\cos 27^\circ }}{{\sin 27^\circ }}} \right)\)

\(\Rightarrow \frac{{{\rm{si}}{{\rm{n}}^2}9^\circ + {\rm{co}}{{\rm{s}}^2}9^\circ }}{{\cos 9{\rm{\;}}^\circ \sin 9^\circ }} - \left( {\frac{{{\rm{si}}{{\rm{n}}^2}27^\circ + {\rm{co}}{{\rm{s}}^2}27^\circ }}{{\cos 27^\circ \sin 27^\circ }}} \right)\)

As we know that, sin² θ + cos² θ = 1

\(\Rightarrow \frac{1}{{\cos 9{\rm{\;}}^\circ \sin 9^\circ }} - \left( {\frac{1}{{\cos 27^\circ \sin 27^\circ }}} \right)\)

\(\Rightarrow \frac{2}{{2{\rm{\;cos}}9^\circ \sin 9^\circ }} - \left( {\frac{2}{{2{\rm{\;cos}}27^\circ \sin 27^\circ }}} \right)\)

As we know that, 2sinθcosθ = sin2θ

\(\Rightarrow \frac{2}{{\sin 18^\circ }} - \left( {\frac{2}{{\sin 54^\circ }}} \right)\)

As we know that, \({\rm{sin}}18^\circ = {\rm{\;}}\frac{{\sqrt 5 - 1}}{4}{\rm{\;and\;sin}}54^\circ = \frac{{\sqrt 5 + 1}}{4}\)

\(\Rightarrow \frac{2}{{\frac{{\sqrt 5 - 1}}{4}}} - \;\frac{2}{{\frac{{\sqrt 5 + 1}}{4}}}\)

\(\Rightarrow \frac{8}{{\sqrt 5 - 1}} - \left( {\frac{8}{{\sqrt 5 + 1}}} \right)\)

\(\Rightarrow \frac{{8\left( {\sqrt 5 + 1} \right) - 8\left( {\sqrt 5 - 1} \right)}}{{\left( {\sqrt 5 - 1} \right)\left( {\sqrt 5 + 1} \right)}}\)

As we know that, (A + B) (A – B) = A² – B²

⇒ 16/4 = 4

CONCEPT:

Trigonometric Identities:

• \({\rm{cosec\theta }} = \frac{1}{{{\rm{sin\theta }}}}\)
• \({\rm{sec\theta }} = \frac{1}{{{\rm{cos\theta }}}}\)
• \({\rm{cos\;}}30^\circ = {\rm{sin\;}}60^\circ = \frac{{\sqrt 3 }}{2}\)
• \({\rm{cos\;}}60^\circ = {\rm{sin\;}}30^\circ = \frac{1}{2}\)
• (cos A× cos B) – (sin A × sin B) = cos(A + B)
• 2 × sin A × cos A = sin (2A)
• sin (90 - θ) = cos θ

 

CALCULATION:

Given that: √3 cosec 20° – sec 20°

As we know that, cosec θ = 1/ sin θ and sec θ = 1/ cos θ

\(\Rightarrow \frac{{\sqrt 3 }}{{\sin 20^\circ }} - \frac{1}{{\cos 20^\circ }}\)

\(\Rightarrow 2\left\{ {\frac{{\frac{{\sqrt 3 }}{2}}}{{\sin 20^\circ }} - \frac{{\frac{1}{2}}}{{\cos 20^\circ }}} \right\}\)

As we know that, cos 30° = √3 / 2, sin 30° = ½

\(\Rightarrow 2\left\{ {\frac{{\cos 30^\circ }}{{\sin 20^\circ }} - \frac{{\sin 30^\circ \;}}{{\cos 20^\circ }}} \right\}\)

\(\Rightarrow 2\left\{ {\frac{{\cos 30^\circ \cos 20^\circ - \sin 30^\circ \sin 20^\circ }}{{\sin 20^\circ \cos 20^\circ }}} \right\}\)

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

\(\Rightarrow 2\left\{ {\frac{{\cos \left( {50^\circ } \right)}}{{\sin 20^\circ \cos 20^\circ }}} \right\}\)

\(\Rightarrow 2 \times 2\;\left\{ {\frac{{\cos \left( {50^\circ } \right)}}{{2 \times \sin 20^\circ \cos 20^\circ }}} \right\}\)

As we know that, 2 × sin A × cos A = sin (2A)

\(\Rightarrow 4\left\{ {\frac{{\cos \left( {50^\circ } \right)}}{{\sin 40^\circ }}} \right\}\)

\(\Rightarrow 4\left\{ {\frac{{\cos \left( {50^\circ } \right)}}{{{\rm{sin}}\left( {90^\circ - 50^\circ } \right)}}} \right\}\)

As we know that, sin (90 - θ) = cos θ

\(\Rightarrow 4\left\{ {\frac{{\cos \left( {50^\circ } \right)}}{{\cos 50^\circ }}} \right\}\)

⇒ 4

CONCEPT:

Trigonometric Identities:

• \({\rm{tan\theta }} = {\rm{\;}}\frac{{{\rm{sin\theta }}}}{{{\rm{cos\theta }}}}\)
• \(\cot {\rm{\theta }} = {\rm{\;}}\frac{{{\rm{cos\theta }}}}{{{\rm{sin\theta }}}}\)
• sin (A + B) + sin (A – B) = 2 sinA cosB

 

Componendo and dividendo rule:

If a : b = c : d then \(\frac{{{\rm{a}} + {\rm{b}}}}{{{\rm{a}} - {\rm{b}}}} = \frac{{{\rm{c}} + {\rm{d}}}}{{{\rm{c}} - {\rm{d}}}}\)

 

CALCULATION:

Given that: A + B = α, A – B = x and tan A : tan B = p : q

Let A + B = α       … equation (1)

Let A – B = x       … equation (2)

By adding equation (1) and equation (2)

⇒ 2A = α + x

⇒ A = (α + x)/2       … equation (3)

By subtracting equation (1) from equation (2)

⇒ 2B = α – x

⇒ B = (α – x)/2       … equation (4)

Now, calculate \(\frac{{\tan A}}{{\tan B}} = \frac{{\tan \left( {\frac{{\alpha + x\;}}{2}} \right)}}{{\tan \left( {\frac{{\alpha - x}}{2}} \right)}}\)

As we know that \({\rm{tan\theta }} = \frac{{{\rm{sin\theta }}}}{{{\rm{cos\theta }}}}{\rm{\;and}}\cot {\rm{\theta }} = {\rm{\;}}\frac{{{\rm{cos\theta }}}}{{{\rm{sin\theta }}}}\)

\(\Rightarrow \frac{{\tan {\rm{A}}}}{{\tan {\rm{B}}}} = \frac{{\sin \left( {\frac{{{\rm{\alpha }} + {\rm{x}}}}{2}} \right)\cos \left( {\frac{{{\rm{\alpha }} - {\rm{x}}}}{2}} \right)}}{{\cos \left( {\frac{{{\rm{\alpha }} + {\rm{x}}}}{2}} \right)\sin \left( {\frac{{{\rm{\alpha }} - {\rm{x}}}}{2}} \right)}}\)

\(\Rightarrow \frac{{\tan {\rm{A}}}}{{\tan {\rm{B}}}} = \frac{{2\sin \left( {\frac{{{\rm{\alpha }} + {\rm{x}}}}{2}} \right)\cos \left( {\frac{{{\rm{\alpha }} - {\rm{x}}}}{2}} \right)}}{{2\cos \left( {\frac{{{\rm{\alpha }} + {\rm{x}}}}{2}} \right)\sin \left( {\frac{{{\rm{\alpha }} - {\rm{x}}}}{2}} \right)}}\)

As we know that sin (A + B) + sin (A – B) = 2 sinA cosB

\(\Rightarrow \frac{{\tan {\rm{A}}}}{{\tan {\rm{B}}}} = \frac{{\sin \left( {\frac{{{\rm{\alpha }} + {\rm{x}} + {\rm{\alpha }} - {\rm{x}}}}{2}} \right) + {\rm{\;sin}}\left( {\frac{{{\rm{\alpha }} + {\rm{x}} - {\rm{\alpha }} + {\rm{x}}}}{2}} \right)}}{{\sin \left( {\frac{{{\rm{\alpha }} + {\rm{x}} + {\rm{\alpha }} - {\rm{x}}}}{2}} \right) + {\rm{\;sin}}\left( {\frac{{{\rm{\alpha }} - {\rm{x}} - {\rm{\alpha }} - {\rm{x}}}}{2}} \right)}}\)

\(\Rightarrow \frac{{\tan {\rm{A}}}}{{\tan {\rm{B}}}} = \frac{{\sin \left( {\rm{\alpha }} \right) + {\rm{\;sin}}\left( {\rm{x}} \right)}}{{\sin \left( {\rm{\alpha }} \right) - {\rm{\;sin}}\left( {\rm{x}} \right)}} = \;\frac{{\rm{p}}}{{\rm{q}}}\)

By componendo & dividendo rule,

\(\Rightarrow \frac{{\sin \left( {\rm{\alpha }} \right) + {\rm{\;sin}}\left( {\rm{x}} \right) + \sin \left( {\rm{\alpha }} \right) - {\rm{\;sin}}\left( {\rm{x}} \right)}}{{\sin \left( {\rm{\alpha }} \right) + {\rm{\;sin}}\left( {\rm{x}} \right) - \sin \left( {\rm{\alpha }} \right) + {\rm{\;sin}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{{{\rm{p}} + {\rm{q}}}}{{{\rm{p}} - {\rm{q}}}}\)

\(\Rightarrow \frac{{{\rm{sin}}\left( {\rm{\alpha }} \right)}}{{{\rm{sin}}\left( {\rm{x}} \right)}} = \frac{{{\rm{p}} + {\rm{q\;}}}}{{{\rm{p}} - {\rm{q}}}}\;\)

\(\Rightarrow \sin {\rm{x}} = \frac{{{\rm{p}} - {\rm{q}}}}{{{\rm{p}} + {\rm{q}}}}{\rm{sin\alpha }}\)

CONCEPT:

Trigonometric functions:

• \({\rm{sin\theta }} = \frac{{{\rm{opposite\;side\;}}}}{{{\rm{hypotenuse}}}}\)
• \({\rm{cos\theta }} = \frac{{{\rm{adjacent\;side}}}}{{{\rm{hypotenuse}}}}\)
• \({\rm{tan\theta }} = \frac{{{\rm{opposite\;side\;}}}}{{{\rm{adjacent\;side}}}}\)

Inverse trigonometric functions:

• \({\rm{ta}}{{\rm{n}}^{ - 1}}{\rm{A}} + {\rm{ta}}{{\rm{n}}^{ - 1}}{\rm{B}} = {\rm{ta}}{{\rm{n}}^{ - 1}}\left\{ {\frac{{{\rm{A}} + {\rm{B\;}}}}{{1 - \left( {{\rm{AB}}} \right)}}} \right\}\)

 

CALCULATION:

Given that: \({\sin ^{ - 1}}\left( {\frac{3}{5}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)\)

Let \({\rm{si}}{{\rm{n}}^{ - 1}}\left( {\frac{3}{5}} \right) = {\rm{\theta }}\)

\(\Rightarrow \sin \left( \theta \right) = \frac{3}{5}\)

As we know that, sin θ = (Opposite side / Hypotenuse)

From Pythagoras’ theorem,

(Hypotenuse)² = (Adjacent side)² + (Opposite side)²

⇒ (5)² = (Adjacent side)² + (3)²

⇒ Adjacent side = 4

As we know that, tan θ = (Opposite side / Adjacent side)

⇒ tan θ = (3/4)

\(\Rightarrow {\rm{\theta }} = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{3}{4}} \right)\)

We had already taken that, \({\rm{si}}{{\rm{n}}^{ - 1}}\left( {\frac{3}{5}} \right) = {\rm{\theta }}\)

So, \({\rm{si}}{{\rm{n}}^{ - 1}}\left( {\frac{3}{5}} \right) = {\rm{\theta }} = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{3}{4}} \right)\)

The given question can be written as: \({\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{3}{4}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)\)

As we know that, \({\rm{ta}}{{\rm{n}}^{ - 1}}{\rm{A}} + {\rm{ta}}{{\rm{n}}^{ - 1}}{\rm{B}} = {\rm{ta}}{{\rm{n}}^{ - 1}}\left\{ {\frac{{{\rm{A}} + {\rm{B\;}}}}{{1 - \left( {{\rm{AB}}} \right)}}} \right\}\)

\(\Rightarrow {\rm{ta}}{{\rm{n}}^{ - 1}}\left\{ {\frac{{\frac{3}{4} + \frac{1}{7}}}{{1 - \left( {\frac{3}{4} \times \frac{1}{7}} \right)}}} \right\}\)

\(\Rightarrow {\rm{ta}}{{\rm{n}}^{ - 1}}\left\{ {\frac{{\frac{{25}}{{21}}}}{{1 - \left( {\frac{3}{{28}}} \right)}}} \right\}\)

As we know that, tan (π /4) = 1

\(\Rightarrow {\rm{ta}}{{\rm{n}}^{ - 1}}\left\{ {\frac{{\frac{{25}}{{28}}}}{{\frac{{25}}{{28}}}}} \right\} = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( 1 \right) = \frac{{\rm{\pi }}}{4}\)

CONCEPT:

Rationalisation:

It is the process by which a fraction is rewritten so that the denominator contains only rational numbers.

Trigonometric Identities:

• \({\rm{tan}}30^\circ = \frac{1}{{\sqrt 3 }}\)
• \({\rm{tan}}45^\circ = 1\)

CALCULATION:

In Δ ABC,

\(\Rightarrow {\rm{tan}}30^\circ = {\rm{\;}}\frac{{{\rm{AB}}}}{{{\rm{BC}}}} = \frac{{{{\rm{h}}_{\rm{t}}} - {{\rm{h}}_{\rm{p}}}}}{{\rm{x}}}\)

\(\Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{{{\rm{h}}_{\rm{t}}} - {{\rm{h}}_{\rm{p}}}}}{{\rm{x}}}\)

⇒ x = √3(h ­_t – h_p)       … equation (1)

In Δ ADE,

\(\Rightarrow {\rm{tan}}45^\circ = \frac{{{\rm{AD}}}}{{{\rm{DE}}}} = \frac{{{{\rm{h}}_{\rm{t}}}}}{{\rm{x}}}\)

⇒ x = h_t­       … equation (2)

Substituting equation (1) in equation (2)

⇒ h _t = √3(h ­_t – h_p)

⇒ √3h_p = h _t (√3 – 1)

\(\Rightarrow \frac{{{{\rm{h}}_{\rm{p}}}}}{{{{\rm{h}}_{\rm{t}}}}} = {\rm{\;}}\frac{{\sqrt 3 - 1}}{{\sqrt 3 }}\)

By rationalisation,

\(\Rightarrow \frac{{{{\rm{h}}_{\rm{p}}}}}{{{{\rm{h}}_{\rm{t}}}}} = {\rm{\;}}\frac{{\sqrt 3 - 1}}{{\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} = \;\frac{{3 - \sqrt 3 }}{3}\)

\(\Rightarrow {{\rm{h}}_{\rm{p}}} = \frac{{3 - \sqrt 3 }}{3}{\rm{\;}}{{\rm{h}}_{\rm{t}}}\)

Hence, options \(\frac{{2{h_P}{h_T}}}{{3 + \sqrt 3 }} = h_P^2,\frac{{{h_T} - {h_P}}}{{\sqrt 3 + 1}} = \frac{{{h_P}}}{2}\) are correct.

CONCEPT:

Properties of Triangles:

If three sides a, b and c are sides of a triangle, \({\rm{\Delta }} = {\rm{\;}}\sqrt {{\rm{s}}\left( {{\rm{s}} - {\rm{a}}} \right)\left( {{\rm{s}} - {\rm{b}}} \right)\left( {{\rm{s}} - {\rm{c}}} \right)} \)where \({\rm{s}} = \frac{{{\rm{a}} + {\rm{b}} + {\rm{c}}}}{2}\) where

S = semi perimeter of a triangle

Half angle formulae of a triangle:

 In Δ ABC

• \(\sin \left( {\frac{{\rm{A}}}{2}} \right) = {\rm{\;}}\sqrt {\frac{{({\rm{s}} - {\rm{b}}(\left( {{\rm{s}} - {\rm{c}}} \right)}}{{{\rm{bc}}}}} \)
• \(\cos \left( {\frac{{\rm{A}}}{2}} \right) = {\rm{\;}}\sqrt {\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{a}}} \right)}}{{{\rm{bc}}}}} \)
• \(\tan \left( {\frac{{\rm{A}}}{2}} \right) = {\rm{\;}}\sqrt {\frac{{\left( {{\rm{s}} - {\rm{b}}} \right)\left( {{\rm{s}} - {\rm{c}}} \right)}}{{{\rm{s}}\left( {{\rm{s}} - {\rm{a}}} \right)}}} \)

 

CALCULATION:

Given that, a – 2b + c = 0

⇒ a + c = 2b

As we know that, S (Semi perimeter) of a triangle = (a + b + c)/2

⇒ s = (2b + b) /2 = 3b/2

As we know that, \({\rm{Cot\;}}\left( {\frac{{\rm{A}}}{2}} \right) = \sqrt {\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{a\;}}} \right)}}{{\left( {{\rm{s}} - {\rm{b}}} \right)\left( {{\rm{s}} - {\rm{c}}} \right)}}} ,\cot \left( {\frac{{\rm{c}}}{2}} \right) = {\rm{\;}}\sqrt {\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{c}}} \right)}}{{\left( {{\rm{s}} - {\rm{a}}} \right)\left( {{\rm{s}} - {\rm{b}}} \right)}}} \)

\(\Rightarrow \cot \left( {\frac{{\rm{A}}}{2}} \right)\cot \left( {\frac{{\rm{c}}}{2}} \right) = {\rm{\;}}\sqrt {\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{a\;}}} \right)}}{{\left( {{\rm{s}} - {\rm{b}}} \right)\left( {{\rm{s}} - {\rm{c}}} \right)}}} \sqrt {\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{c}}} \right)}}{{\left( {{\rm{s}} - {\rm{a}}} \right)\left( {{\rm{s}} - {\rm{b}}} \right)}}} \)

\(\Rightarrow \cot \left( {\frac{{\rm{A}}}{2}} \right)\cot \left( {\frac{{\rm{c}}}{2}} \right) = {\rm{\;}}\sqrt {\frac{{{{\rm{s}}^2}}}{{{{\left( {{\rm{s}} - {\rm{b}}} \right)}^2}}}} \)

\(\Rightarrow \cot \left( {\frac{{\rm{A}}}{2}} \right)\cot \left( {\frac{{\rm{c}}}{2}} \right) = \frac{{\rm{s}}}{{\left( {{\rm{s}} - {\rm{b}}} \right)}}\)

\(\Rightarrow \cot \left( {\frac{{\rm{A}}}{2}} \right)\cot \left( {\frac{{\rm{c}}}{2}} \right) = \frac{{\frac{{3{\rm{b}}}}{2}}}{{\frac{{3{\rm{b}}}}{2} - {\rm{b}}}} = 3\)

CALCULATION:

Given that: \(\sqrt {1 + \sin A} = - \left( {\sin \frac{A}{2} + \cos \frac{A}{2}} \right)\)

\(\Rightarrow \sqrt {1 + \sin A} = \sqrt {{{\sin }^2}A + {{\cos }^2}A + 2\cos \frac{{\rm{A}}}{2}\sin \frac{A}{2}} = \left| {\sin \frac{A}{2} + \cos \frac{A}{2}} \right|\)

If \(\frac{{3{\rm{\pi }}}}{2} < {\rm{A}} < \frac{{7{\rm{\pi }}}}{2}\)

Than,

\(\Rightarrow \left| {\sin \frac{A}{2} + \cos \frac{A}{2}} \right| = - \left( {\sin \frac{A}{2} + \cos \frac{A}{2}} \right)\)

Hence Option 3 is correct.

CONCEPT:

Trigonometric Identities:

• sin² A + cos² A = 1
• cos (A + B) = cos A cos B – sin A sin B
• cos² A – sin² B = cos(A+B) cos (A – B)
• cos (π – A) = cos A

Sum of interior angles of a triangle = 180° = π

 

CALCULATION:

Given that,

Solution: \(\frac{{{{\sin }^2}A + {{\sin }^2}B + {{\sin }^2}C}}{{{{\cos }^2}A + {{\cos }^2}B + {{\cos }^2}C}} = 2\)

⇒ sin² A + sin² B + sin² c = 2(cos² A + cos² B + cos² c)

As we know that, sin² A + cos² A = 1

⇒ sin² A + sin² B + sin² c = 2{(1- sin² A) + (1 – sin² B) + (1 – sin² c)}

⇒ sin² A + sin² B + sin² c = 2{3 – sin² A – sin² B – sin² B}

⇒ 3(sin² A + sin² B + sin² c) = 6

⇒ sin² A + sin² B + sin² c = 2

As we know that, sin² A + cos² A = 1

⇒ (1 – cos² A) +(1 – cos² B) + (1 – cos² C) = 2

⇒ cos²A + cos² B + cos² C = 1

⇒ cos²A + cos² B = 1 – cos² C

⇒ cos²A + cos² B = sin² C

⇒ cos²A + cos² B – sin² C = 0

As we know that, cos² A – sin² B = cos(A+B) cos (A – B)

⇒ cos²A + cos (B + C) cos (B – C) = 0

As we know that, sum of interior angles of a triangle = A + B + C = 180°

⇒ cos²A + cos (π – A) cos (B – C) = 0

As we know that, cos (π – A) = cos A

⇒ cos²A + cos A cos (B – C) = 0

⇒ cos A (cos A – cos (B – c)) = 0

As we know that, cos (π – A) = cos A

⇒ cos A (cos (π – (B + C)) – cos (B – C) = 0

⇒ cos A (cos (B + C) – cos (B – C)) = 0

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

⇒ cos A (cos B cos C – sin B sin C – (cos B cos C + sin B sin C)) = 0

⇒ cos A (2 cos B cos C) = 0

⇒ cos A cos B cos C = 0

The above condition is possible only when either cos A = 0 or cos B = 0 or cos C = 0

As we know that, cos (π / 2) = 0

⇒ ∠ A = π /2 or ∠ B = π /2 or ∠ C = π / 2

Hence, the given triangle is a right - angled triangle.

CONCEPT:

Inverse sine function:

The function sin^-1: [ -1, 1] → \(\left[ {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right]\) defined by sin^-1 x = θ ⇔ θ ∈ \(\left[ {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right]\) and sin θ = x, for all x ∈ [-1, 1]

 

CALCULATION:

Let us assume sin^-1 x = y

⇒ x = sin y

We know that, maximum and minimum values of sine function lie between -1 to 1

⇒ -1 ≤ sin y ≤ 1

⇒ -1 ≤ x ≤ 1

Apply sin^-1 for the above inequality

⇒ sin^-1 (-1) ≤ sin^-1 x ≤ sin^-1 (1)

⇒ (-π /2) ≤ sin^-1 x ≤ (π /2)

Hence, the principal value of sin^-1 x lies in the interval \(\left[ {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right]\)