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Concept:

• sin (2nπ ± θ) = ±  sin θ, where n ∈ Z.
• cos (2nπ ± θ) = cos θ , where n ∈ Z.
• cos (-θ) = cos θ
• cos (2nπ - θ) = cos θ

Calculation:

As we know that, the value of cos θ repeats after an interval of 2π or 360°.

⇒ cos (-1740°) = cos (1740°)               (∵ cos (-θ) = cos θ)

We know that cos (2nπ - θ) = cos θ 

cos (1740°)= cos (5 × 360° - 1740°) = cos (1800° - 1740°) = cos (60°)

As we know that, cos 60° = 1 / 2.

⇒ cos (-1740°) = 1 / 2.

Given:

Set A = {1, 2, 3, 4, 5}

Concept:

Subset - Set A is said to be a subset of Set B if all the elements of Set A are also present in Set B.

Solution:

A subset of A which contains element 2 but not 4 as follows,

{1, 2, 3, 5}, {1, 2, 5}, {1, 2, 3}, {1, 2}, {2}, {2, 3}, {2, 5}, {2, 3, 5}

The total numbers of the subset are 8.

Concept

\(\tan (x-y) = \frac{\tan x - \tan y}{1+\tan x\tan y}\)

Calculation:

We have to find the value of tan 15°

⇒ tan 15° = tan (45° -  30°)

 \( = \frac{\tan 45 - \tan 30}{1+\tan 45\tan 30}\)                   [∵ \(\tan (x-y) = \frac{\tan x - \tan y}{1+\tan x\tan y}\)]

\(=\frac{1 - \frac{1}{\sqrt3}}{1+1\times\frac{1}{\sqrt3}}\)

\(=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}\)

\(=\frac{\sqrt{3}-1}{\sqrt{3}+1}\)

\(=\frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}\)

\(=\frac{(\sqrt{3}-1)^2}{(\sqrt{3})^2-1^2} \)

\(=\frac{3+1-2\sqrt3}{3-1}= \frac{4-2\sqrt3}{2}=2-\sqrt3\)

We have n = 14, then the number of terms would be 15.

\(\therefore \left( {\frac{{14}}{2} + 1} \right)\) i.e 8^th will be middle term

\(a = 1,b = \frac{{{x^2}}}{2},n = 14,r = 7\)

\({T_{r + 1}} = {}_;^n{C_r}{a^{n - r}}{b^r}\)

\({T_{7 + 1}} = {}_;^{14}{C_7}{\left( 1 \right)^{14 - 7}}{\left( { - \frac{{{x^2}}}{2}} \right)^7}\)

\( = \frac{{14!}}{{7!7!}}{\left( { - 1} \right)^7}\frac{{{x^{14}}}}{{{2^7}}}\)

\( = \frac{{14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7!}}{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 7!}}\left( {\frac{{\left( { - 1} \right)}}{{128}}{x^{14}}} \right)\)

\( = - \frac{{429}}{{16}}{x^{14}}\)

Given:

We have to evalute 2 tan2 45° + cos2 30° - sin2 60°

Formula Used:

tan45° = 1 

Cos30° = √3/2

Sin60° = √3/2

Calculation:

2 tan² 45° + cos² 30° - sin² 60°

⇒ 2 × 1² + (√3/2)² – (√3/2)²

⇒ 2 + 3/4 – 3/4

⇒ 2

∴ The value of 2 tan2 45° + cos2 30° - sin2 60° is 2.

Concept:

• Let, (a ± ib) be a complex number then the square root of a ± ib is given by

\(\sqrt {{\rm{a}} + {\rm{ib}}} = \pm \left( {\sqrt {\frac{{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} + {\rm{a}}}}{2}} \pm i{\rm{;}}\sqrt {\frac{{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} - {\rm{a}}}}{2}} } \right)\)

Calculation:

Here, 4 + 3i

Comparing with standard form a + bi, we get

a = 4, b = 3

\(\sqrt {{\rm{a}} + {\rm{ib}}} = \pm \left( {\sqrt {\frac{{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} + {\rm{a}}}}{2}} + {\rm{i;}}\sqrt {\frac{{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} - {\rm{a}}}}{2}} } \right)\)

\(\therefore \sqrt {4 + 3{\rm{i}}} = {\rm{;}} \pm \left( {\sqrt {\frac{{\sqrt {{4^2} + {3^2}} + 4}}{2}} + {\rm{i}}\sqrt {\frac{{\sqrt {{4^2} + {3^2}} - 4}}{2}} } \right)\)

\(\Rightarrow \pm \left( {\sqrt {\frac{{\sqrt {25} + 4}}{2} }\ + \ {\rm{i}}\sqrt {\frac{{\sqrt {25} - 4}}{2}} } \right)\)

\(\Rightarrow \pm \left( {\sqrt {\frac{{5 + 4}}{2}}\ + \ {\rm{i}}\sqrt {\frac{{5 - 4}}{2}} } \right) = \pm \left( {\sqrt {\frac{9}{2}} \ +\ {\rm{i}}\sqrt {\frac{1}{2}} } \right)\)

\(\therefore \sqrt {4 + 3{\rm{i}}} = \pm \left( {\frac{3}{{\sqrt 2 }} + {\rm{i}}\frac{1}{{\sqrt 2 }}} \right)\)

Hence, option (4) is correct.

Concept: 

Let z = x + iy be a complex number.

• Modulus of z = \(\left| {\rm{z}} \right| = {\rm{}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} = {\rm{}}\sqrt {{\rm{Re}}{{\left( {\rm{z}} \right)}^2} + {\rm{Im;}}{{\left( {\rm{z}} \right)}^2}}\)
• arg (z) = arg (x + iy) = \({\tan ^{ - 1}}\left( {\frac{y}{x}} \right)\)
• Conjugate of z = z̅ = x – iy

Calculation:

Given complex number is z = \(\rm 3i+4\over2-3i\)

z = \(\rm {3i+4\over2-3i}\times{2+3i\over2+3i}\)

z = \(\rm 6i+8-9+12i\over2^2-(3i)^2\)

z = \(\rm 18i-1\over13\)

z = \(\rm {-1\over13}+{18\over13}i\)

Conjugate of z = (z̅) = \(\rm {-1\over13}-{18\over13}i\)

Calculation:

Given that

A = {2, 3, 4}

B = {3, 5}

Union of A and B 

A ∪ B = {2, 3, 4, 5}

Therefore, A ∪ B = {2, 3, 4, 5}.

Calculation:

(n+2) ! = 2550 × n !

(n + 2).(n + 1).n ! = 2550 n!

(n + 2).(n + 1) = 51 × 50

On comparing both sides

n + 1 = 50

∴ n = 49

The correct answer is 3780

Explanation:

• Number of vowels = 5
• Number of consonant = 21

Here, our condition is to take atleast one consonant, but not consecutive, so we are left with two cases

Case I: Only one consonant

Number of ways 3 × 21 × 5 × 5 = 1575

= 21 × 5 × 4 × 3 = 1260

Case II: Two consonants, but at alternate positions,

Number of ways = 21 × 21 × 5 = 2205

Therefore, total number of ways = 1575 + 2205 = 3780