JEE Main Previous Year Trigonometry Questions and Solutions

Q 1 :

If $\sin x = - \frac{3}{5},$ where $π < x < \frac{3 π}{2},$ then $80 (\tan^{2} x - \cos x)$ is equal to [2024]

109
108
19
18

1

2

3

4

(1)

$\sin x = \frac{- 3}{5},$ $π < x < \frac{3 π}{2}$

$\cos x = \sqrt{1 - \sin^{2} x} = \sqrt{1 - \frac{9}{25}} = \frac{- 4}{5}$ $[$ $∵$ $x$ lies in third quadrant $]$

Now, $\tan x = \frac{3}{4}$

So, $80 (\tan^{2} x - \cos x) = 80 (\frac{9}{16} + \frac{4}{5}) = 45 + 64 = 109$

Q 2 :

If the value of $\frac{3 \cos 36^{o} + 5 \sin 18^{o}}{5 \cos 36^{o} - 3 \sin 18^{o}}$ is $\frac{a \sqrt{5} - b}{c},$ where $a, b, c$ are natural numbers and $g c d (a, c) = 1,$ then $a + b + c$ is equal to : [2024]

52
50
54
40

1

2

3

4

(1)

$\frac{3 \cos 36^{o} + 5 \sin 18^{o}}{5 \cos 36^{o} - 3 \sin 18^{o}}$ $= \frac{3 [\frac{1 + \sqrt{5}}{4}] + 5 [\frac{\sqrt{5} - 1}{4}]}{5 [\frac{1 + \sqrt{5}}{4}] - 3 [\frac{\sqrt{5} - 1}{4}]}$ $[∵ \sin 18^{o} = \frac{\sqrt{5} - 1}{4} and \cos 36^{o} = \frac{1 + \sqrt{5}}{4}]$

$= \frac{3 + 3 \sqrt{5} + 5 \sqrt{5} - 5}{5 + 5 \sqrt{5} - 3 \sqrt{5} + 3} = \frac{8 \sqrt{5} - 2}{2 \sqrt{5} + 8} = \frac{4 \sqrt{5} - 1}{4 + \sqrt{5}}$

$= \frac{(4 \sqrt{5} - 1) (4 - \sqrt{5})}{(\sqrt{5} + 4) (4 - \sqrt{5})} = \frac{16 \sqrt{5} - 20 - 4 + \sqrt{5}}{16 - 5}$

$= \frac{17 \sqrt{5} - 24}{11} = \frac{a \sqrt{5} - b}{c}$ (Given)

$∴$ $a = 17,$ $b = 24,$ $c = 11$

$∴$ $a + b + c = 17 + 24 + 11 = 52$

Q 3 :

If $\tan A = \frac{1}{\sqrt{x (x^{2} + x + 1)}},$ $\tan B = \frac{\sqrt{x}}{\sqrt{x^{2} + x + 1}}$ and $\tan C = {(x^{- 3} + x^{- 2} + x^{- 1})}^{1 / 2}, 0 < A, B, C < π / 2,$ then $A + B$ is equal to: [2024]

$π - C$
$2 π - C$
$C$
$\frac{π}{2} - C$

1

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(3)

As $\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

$= \frac{\frac{1}{\sqrt{x (x^{2} + x + 1)}} + \frac{\sqrt{x}}{\sqrt{x^{2} + x + 1}}}{1 - \frac{1}{x^{2} + x + 1}}$

$= \frac{1 + x}{\sqrt{x (x^{2} + x + 1)}} \times \frac{x^{2} + x + 1}{x^{2} + x} = \frac{1 + x}{\sqrt{x}} \times \frac{\sqrt{x^{2} + x + 1}}{x (1 + x)}$

$= \sqrt{\frac{x^{2} + x + 1}{x^{3}}} = \sqrt{x^{- 1} + x^{- 2} + x^{- 3}} = \tan C$

$\Rightarrow A + B = C$

Q 4 :

Let $A = {θ \in [0, 2 π] : 1 + 10 Re (\frac{2 \cos θ + i \sin θ}{\cos θ - 3 i \sin θ}) = 0}$ .

Then $\sum_{θ \in A} θ^{2}$ is equal to [2025]

$\frac{27}{4} π^{2}$
$6 π^{2}$
$\frac{21}{4} π^{2}$
$8 π^{2}$

1

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4

(3)

We have,

$\frac{2 \cos θ + i \sin θ}{\cos θ - 3 i \sin θ} = \frac{2 \cos θ + i \sin θ}{\cos θ - 3 i \sin θ} \times \frac{\cos θ + 3 i \sin θ}{\cos θ + 3 i \sin θ}$

$= \frac{2 \cos^{2} θ + 6 i \cos θ \sin θ + i \sin θ \cos θ - 3 \sin^{2} θ}{\cos^{2} θ + 9 \sin^{2} θ}$

$= \frac{2 \cos^{2} θ - 3 \sin^{2} θ}{\cos^{2} θ + 9 \sin^{2} θ} + \frac{i (7 \cos θ \sin θ)}{\cos^{2} θ + 9 \sin^{2} θ}$

Now, $1 + 10 Re (\frac{2 \cos θ + i \sin θ}{\cos θ - 3 i \sin θ}) = 0$

$\Rightarrow 1 + \frac{20 \cos^{2} θ - 30 \sin^{2} θ}{\cos^{2} θ + 9 \sin^{2} θ} = 0$

$\Rightarrow \cos^{2} θ = \sin^{2} θ$

$\Rightarrow \tan^{2} θ = 1 \Rightarrow \tan θ = \pm 1$

$\Rightarrow θ = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$

$∴ \sum_{θ \in A} θ^{2} = \frac{π^{2}}{16} + \frac{9 π^{2}}{16} + \frac{25 π^{2}}{16} + \frac{49 π^{2}}{16} = \frac{84 π^{2}}{16} = \frac{21 π^{2}}{4}$

Q 5 :

The value of (sin 70°) (cot 10° cot 70° – 1) is [2025]

0
1
3/2
2/3

1

2

3

4

(2)

We have,

$\sin 70 ° (\frac{\cos 10 ° \cos 70 ° - \sin 10 ° \sin 70 °}{\sin 10 ° \sin 70 °})$

$= \sin 70 ° [\frac{\cos 80 °}{\sin 10 ° \sin 70 °}]$ [ $∵$ cos (A + B) = cos A cos B – sin A sin B]

$= \frac{\cos 80 °}{\cos 80 °} = 1$ [ $∵ \sin (90 ° - θ) = \cos θ$ ]

Q 6 :

If $\sum_{r = 1}^{13} {\frac{1}{\sin (\frac{π}{4} + (r - 1) \frac{π}{6}) \sin (\frac{π}{4} + \frac{r π}{6})}} = a \sqrt{3} + b, b \in Z,$ then $a^{2} + b^{2}$ is equal to : [2025]

2
4
10
8

1

2

3

4

(4)

The given expression can be written as,

$\frac{1}{\sin \frac{π}{6}} \sum_{r = 1}^{13} \frac{\sin [(\frac{π}{4} + \frac{r π}{6}) - [(\frac{π}{4}) + (r - 1) \frac{π}{6}]]}{\sin (\frac{π}{4} + (r - 1) \frac{π}{6}) \sin (\frac{π}{4} + \frac{r π}{6})}$

$= \frac{1}{\sin \frac{π}{6}} \sum_{r = 1}^{13} (\cot (\frac{π}{4} + (r - 1) \frac{π}{6}) - \cot (\frac{π}{4} + \frac{rπ}{6}))$

$[∵ \frac{\sin (A - B)}{\sin A \sin B} = \cot B - \cot A]$

$= 2 [\cot \frac{π}{4} - \cot (\frac{π}{4} + \frac{13 π}{6})] = 2 (1 - 2 + \sqrt{3})$

$= 2 \sqrt{3} - 2 = a \sqrt{3} + b$ [Given]

$∴ a = 2, b = - 2$

So, $a^{2} + b^{2} = 8$

Q 7 :

Let the line x + y = 1 meet the axes of x and y at A and B, respectively. A right angled triangle AMN is inscribed in the triangle OAB, where O is the origin and the points M and N lie on the lines OB and AB, respectively. If the area of the triangle AMN is $\frac{4}{9}$ of the area of the triangle OAB and AN : NB = $λ$ : 1, then the sum of all possible value(s) of $λ$ is : [2025]

$\frac{5}{2}$
$\frac{13}{6}$
$\frac{1}{2}$
2

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(4)

Area of $△ O A B = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$

Now, Area of $△$ AMN = $\frac{4}{9}$ Area of $△$ OAB

$= \frac{4}{9} \times \frac{1}{2} = \frac{2}{9}$ ... (i)

Since, $∠$ OAB = 45°, then let $∠$ MAN = $θ$ and $∠$ MAO = 45° – $θ$ , then AM = sec (45° – $θ$ );

$A N = A M \times \frac{A N}{A M} = \sec (45 ° - θ) \cos θ$

and $M N = A M \times \frac{M N}{A M} = \sec (45 ° - θ) \sin θ$

Now, $Area of △ A M N = \frac{1}{2} \times A N \times M N$

$= \frac{1}{2} \times \sec^{2} (45 ° - θ) \sin θ \cos θ = \frac{2}{9} [From (i)]$

$\Rightarrow \frac{\sin 2 θ}{1 + \sin 2 θ} = \frac{4}{9} \Rightarrow \sin 2 θ = \frac{4}{5}$

$\Rightarrow 2 \tan^{2} θ - 5 \tan θ + 2 = 0$

$\Rightarrow \tan θ = \frac{1}{2}, 2 (Rejected as θ < 45 °)$

$∵ ∠ O B A = 45 °$

$∴ \tan 45 ° = \frac{M N}{B N} \Rightarrow M N = B N$

Now, $In △ A M N, \cot θ = \frac{A N}{M N} = \frac{A N}{B N} = \frac{λ}{1} \Rightarrow λ = 2$ $[∵ \tan θ = \frac{1}{2}]$

Hence, required sum is 2.

Q 8 :

If $\sin x + \sin^{2} x = 1, x \in (0, \frac{π}{2})$ , then $(\cos^{12} x + \tan^{12} x) + 3 (\cos^{10} x + \tan^{10} x + \cos^{8} x + \tan^{8} x) + (\cos^{6} x + \tan^{6} x)$ is equal to : [2025]

2
4
3
1

1

2

3

4

(1)

We have, $\sin x + \sin^{2} x = 1$

$\Rightarrow \sin x = 1 - \sin^{2} x \Rightarrow \sin x = \cos^{2} x \Rightarrow \tan x = \cos x$

Now, $(\cos^{12} x + \tan^{12} x) + 3 (\cos^{10} x + \tan^{10} x + \cos^{8} x + \tan^{8} x) + (\cos^{6} x + \tan^{6} x)$

$= (\cos^{12} x + \cos^{12} x) + 3 (\cos^{10} x + \cos^{10} x + \cos^{8} x + \cos^{8} x) + (\cos^{6} x + \cos^{6} x)$

$= 2 \cos^{12} x + 6 \cos^{10} x + 6 \cos^{8} x + 2 \cos^{6} x$

$= 2 \sin^{6} x + 6 \sin^{5} x + 6 \sin^{4} x + 2 \sin^{3} x$ [ $∵ \sin x = \cos^{2} x$ ]

$= 2 \sin^{3} x [\sin^{3} x + 3 \sin^{2} x + 3 \sin x + 1]$

$= 2 \sin^{3} x [{(1 + \sin x)}^{3}]$

$= 2 [{(\sin x + \sin^{2} x)}^{3}] = 2$ { $∵ \sin x + \sin^{2} x = 1$ }

Q 9 :

The value of $36 (4 \cos^{2} 9 ° - 1) (4 \cos^{2} 27 ° - 1) (4 \cos^{2} 81 ° - 1) (4 \cos^{2} 243 ° - 1)$ is [2023]

18
54
27
36

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3

4

(4)

We have, $4 \cos^{2} θ - 1 = 4 (1 - \sin^{2} θ) - 1 = 3 - 4 \sin^{2} θ = \frac{\sin 3 θ}{\sin θ}$

$So, given expression can be written as,$

$= 36 \times \frac{\sin 27 °}{\sin 9 °} \times \frac{\sin 81 °}{\sin 27 °} \times \frac{\sin 243 °}{\sin 81 °} \times \frac{\sin 729 °}{\sin 243 °} = 36 \times \frac{\sin 729 °}{\sin 9 °} = 36$

Q 10 :

$96 \cos \frac{π}{33} \cos \frac{2 π}{33} \cos \frac{4 π}{33} \cos \frac{8 π}{33} \cos \frac{16 π}{33}$ is equal to [2023]

3
2
1
4

1

2

3

4

(1)

We know that,

$\cos \frac{π}{33} \cos \frac{2 π}{33} \cos \frac{4 π}{33} \cos \frac{8 π}{33} \cos \frac{16 π}{33} = \frac{\sin \frac{32 π}{33}}{32 \sin \frac{π}{33}}$

$\frac{\sin (π - \frac{π}{33})}{32 \cdot \sin \frac{π}{33}} = \frac{\sin \frac{π}{33}}{32 \cdot \sin \frac{π}{33}} = \frac{1}{32}$

$∴$ $96 \cos \frac{π}{33} \cos \frac{2 π}{33} \cos \frac{4 π}{33} \cos \frac{8 π}{33} \cos \frac{16 π}{33} = \frac{96}{32} = 3$

Q 11 :

The value of $\tan 9 ° - \tan 27 ° - \tan 63 ° + \tan 81 °$ is ________ . [2023]

(4)

$We have, \tan 9 ° - \tan 27 ° - \tan 63 ° + \tan 81 °$

$= \tan 9 ° - \tan 27 ° - \tan (90 ° - 27 °) + \tan (90 ° - 9 °)$

$= (\tan 9 ° + \cot 9 °) - (\tan 27 ° + \cot 27 °)$

$= \frac{2 (\sin^{2} 9 ° + \cos^{2} 9 °)}{2 (\sin 9 ° \cos 9 °)} - \frac{2 (\sin^{2} 27 ° + \cos^{2} 27 °)}{2 (\sin 27 ° \cos 27 °)}$

$= \frac{2}{\sin 18 °} - \frac{2}{\sin 54 °}$

$= \frac{8}{\sqrt{5} - 1} - \frac{8}{\sqrt{5} + 1}$

$= \frac{16}{4} = 4$

Q 12 :

Let $α$ and $β$ respectively be the maximum and the minimum values of the function

$f (θ) = 4 (\sin^{4} (\frac{7 π}{2} - θ) + \sin^{4} (11 π + θ)) - 2 (\sin^{6} (\frac{3 π}{2} - θ) + \sin^{6} (9 π - θ)), θ \in R .$

Then $α + 2 β$ is equal to: [2026]

5
3
4
6

1

2

3

4

(1)

$f (θ) = 4 (\sin^{4} (\frac{7 π}{2} - θ) + \sin^{4} (11 π + θ)) - 2 (\sin^{6} (\frac{3 π}{2} - θ) + \sin^{6} (9 π - θ))$

$f (θ) = 4 (\cos^{4} θ + \sin^{4} θ) - 2 (\cos^{6} θ + \sin^{6} θ)$

$f (θ) = 4 (1 - 2 \sin^{2} θ \cos^{2} θ) - 2 (1 - 3 \sin^{2} θ \cos^{2} θ)$

$f (θ) = 2 - 2 \sin^{2} θ \cos^{2} θ$

$f (θ) = 2 - \frac{\sin^{2} (2 θ)}{2}$

$α = f {(θ)}_{\max} = 2$

$β = f {(θ)}_{\min} = \frac{3}{2}$

$\Rightarrow α + 2 β = 5$

$Ans. = 5 option (1)$

Q 13 :

The value of cosec $10^{o} - \sqrt{3}$ sec 10° is equal to [2026]

2
4
6
8

1

2

3

4

(2)

$= \frac{1}{\sin 10 °} - \frac{\sqrt{3}}{\cos 10 °}$

$= \frac{\cos 10 ° - \sqrt{3} \sin 10 °}{\sin 10 ° \cos 10 °}$

$= 4 [\frac{\frac{1}{2} \cos 10 ° - \frac{\sqrt{3}}{2} \sin 10 °}{2 \sin 10 ° \cos 10 °}]$

$= 4 [\frac{\sin (30 ° - 10 °)}{\sin 20 °}]$

$= 4$

Q 14 :

If $\cot x = \frac{5}{12}$ for some $x \in (π, \frac{3 π}{2}),$ then $\sin 7 x (\cos \frac{13 x}{2} + \sin \frac{13 x}{2}) + \cos 7 x (\cos \frac{13 x}{2} - \sin \frac{13 x}{2})$ is equal to [2026]

$\frac{1}{\sqrt{13}}$
$\frac{6}{\sqrt{26}}$
$\frac{5}{\sqrt{13}}$
$\frac{4}{\sqrt{26}}$

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(1)

$\cot x = \frac{5}{12} \Rightarrow \cos x = - \frac{5}{13} = 2 \cos^{2} \frac{x}{2} - 1$

$\cos (\frac{x}{2}) = - \frac{2}{\sqrt{13}} or \frac{2}{\sqrt{13}} (rejected)$

${∵ \frac{x}{2} \in (\frac{π}{2}, \frac{3 π}{4})}$

$(\sin 7 x \cdot \frac{\sin 13 x}{2} + \cos 7 x \cdot \frac{\cos 13 x}{2}) + (\sin 7 x \cdot \frac{\cos 13 x}{2} - \cos 7 x \cdot \frac{\sin 13 x}{2})$

$= \cos (7 x - \frac{13 x}{2}) + \sin (7 x - \frac{13 x}{2})$

$= \cos \frac{x}{2} + \sin \frac{x}{2}$

$= - \frac{2}{\sqrt{13}} + \frac{3}{\sqrt{13}} = \frac{1}{\sqrt{13}}$

Q 15 :

The value of $\frac{\sqrt{3} c o s e c 20 ° - \sec 20 °}{\cos 20 ° \cos 40 ° \cos 60 ° \cos 80 °}$ is equal to: [2026]

64
16
12
32

1

2

3

4

(1)

$E = \frac{\frac{\sqrt{3}}{\sin 20 °} - \frac{1}{\cos 20 °}}{\frac{1}{2} \cdot \frac{1}{4} \cdot \cos 60 °}$

$= \frac{\sqrt{3} \cos 20 ° - \sin 20 °}{\cos 20 ° . \sin 20 °} \cdot 16$

$= \frac{(\frac{\sqrt{3}}{2} \cos 20 ° - \frac{1}{2} \sin 20 °) 32 \times 2}{2 \cos 20 ° . \sin 20 °}$

$= \frac{\sin 40 °}{\sin 40 °} \times 64 = 64$

Q 16 :

If $\frac{\cos^{2} 48 ° - \sin^{2} 12 °}{\sin^{2} 24 ° - \sin^{2} 6 °} = \frac{α + β \sqrt{5}}{2},$ where $α, β \in ℕ$ , then $α + β$ is equal to __________ . [2026]

(4)

$Use \sin (A + B) \sin (A - B) = \sin^{2} A - \sin^{2} B$

$\cos (A + B) \cos (A - B) = \cos^{2} A - \sin^{2} B$

$\frac{\cos 60 ° \cos 36 °}{\sin 30 ° \sin 18 °} = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} \times \frac{\sqrt{5} + 1}{\sqrt{5} + 1} = \frac{{(\sqrt{5} + 1)}^{2}}{4}$

$= \frac{3 + \sqrt{5}}{2}$

$α = 3, β = 1$

$So, (α + β) = 4$

Q 17 :

The number of elements in the set ${x \in [0, 180 °] : \tan (x + 100 °) = \tan (x + 50 °) \tan x \tan (x - 50 °)}$ is __________ . [2026]

(4)

$\frac{\tan (x + 100 °)}{\tan x} = \tan (x + 50 °) \tan (x - 50 °)$

$\frac{\sin (x + 100 °) \cos x}{\cos (x + 100 °) \sin x} = \frac{\sin (x + 50 °) \sin (x - 50 °)}{\cos (x + 50 °) \cos (x - 50 °)}$

$Apply C & D$

$\frac{\sin (2 x + 100 °)}{\sin 100 °} = \frac{\cos 100 °}{- \cos 2 x}$

$2 \sin (2 x + 100 °) \cos 2 x + \sin 200 ° = 0$

$\sin (4 x + 100 °) + \sin 100 ° + \sin 200 ° = 0$

$\sin (4 x + 100 °) = - 2 \sin 150 ° \cos 50 °$

$\sin (4 x + 100 °) = - \cos 50 ° = \sin (- 40 °)$

$∴$ $4 x + 100 ° = n π + {(- 1)}^{n} \cdot (- 40 °)$

$x = \frac{n π + {(- 1)}^{n + 1} (40 °) - 100 °}{4}$

$∴$ $x = 30 °, 55 °, 120 °, 145 ° in (0, π)$

$∴$ $no. of solutions = 4$

Q 18 :

Let $\frac{π}{2} < θ < π$ and $\cot θ = - \frac{1}{2 \sqrt{2}}$ . Then the value of $\sin (\frac{15 θ}{2}) (\cos 8 θ + \sin 8 θ) + \cos (\frac{15 θ}{2}) (\cos 8 θ - \sin 8 θ)$ is equal to [2026]

$\frac{\sqrt{2} - 1}{\sqrt{3}}$
$\frac{1 - \sqrt{2}}{\sqrt{3}}$
$- \frac{\sqrt{2}}{\sqrt{3}}$
$\frac{\sqrt{2}}{\sqrt{3}}$

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(2)

$\frac{π}{2} < θ < π and \cot θ = - \frac{1}{2 \sqrt{2}}$

$\Rightarrow \sin (\frac{15 θ}{2}) (\cos 8 θ + \sin 8 θ) + \cos (\frac{15 θ}{2}) (\cos 8 θ - \sin 8 θ)$

$\Rightarrow \sin (\frac{15 θ}{2}) \cos 8 θ - \cos (\frac{15 θ}{2}) \sin 8 θ + \sin (\frac{15 θ}{2}) \sin 8 θ + \cos (\frac{15 θ}{2}) \cos 8 θ$

$\Rightarrow \sin (\frac{15 θ}{2} - 8 θ) + \cos (\frac{15 θ}{2} - 8 θ)$

$\Rightarrow \cos \frac{θ}{2} - \sin \frac{θ}{2} = - \sqrt{1 - \sin θ} (\frac{π}{4} < \frac{θ}{2} < \frac{π}{2})$

$\Rightarrow given \cot θ = - \frac{1}{2 \sqrt{2}}, \sin θ = \frac{2 \sqrt{2}}{3}$

$\Rightarrow - \sqrt{1 - \sin θ} = \sqrt{\frac{3 - 2 \sqrt{2}}{3}} = \frac{(\sqrt{2} - 1)}{3}$

$= \frac{1 - \sqrt{2}}{\sqrt{3}}$

Q 19 :

Let $\cos (α + β) = - \frac{1}{10}$ and $\sin (α - β) = \frac{3}{8}$ , where $0 < α < \frac{π}{3}$ and $0 < β < \frac{π}{4}$ , If $\tan 2 α = \frac{3 (1 - r \sqrt{5})}{\sqrt{11} (s + \sqrt{5})}, r, s \in ℕ,$ then r + s is equal to______. [2026]

(20)

$\tan 2 α = \tan [(α + β) + (α - β)]$

$\tan 2 α = \frac{\tan (α + β) + \tan (α - β)}{1 - \tan (α + β) \tan (α - β)}$

$\tan 2 α = \frac{- \sqrt{99} + \frac{3}{\sqrt{55}}}{1 - (\sqrt{99}) (\frac{3}{\sqrt{55}})}$

$\tan 2 α = \frac{- 3 \sqrt{11} + \frac{3}{\sqrt{5} \times \sqrt{11}}}{1 + \frac{9 \sqrt{11}}{\sqrt{5} \times \sqrt{11}}}$

$\tan 2 α = \frac{3 (1 - 11 \sqrt{5})}{\sqrt{11} (9 + \sqrt{5})}$

$r = 11, s = 9$

$r + s = 20$

Q 20 :

If $\frac{\tan (A - B)}{\tan A} + \frac{\sin^{2} C}{\sin^{2} A} = 1, A, B, C \in ((0, \frac{π}{2}))$ , then [2026]

tanA,tanC,tanB are in G.P.
tanA,tanB,tanC are in A.P.
tanA,tanB,tanC are in G.P.
tanA,tanC,tanB are in A.P.

1

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(1)

$\frac{\tan A - \tan B}{(1 + \tan A \tan B) \tan A} + \frac{1 + \cot^{2} A}{1 + \cot^{2} C} = 1$

$Put \tan A = x, \tan B = y, \tan C = z$

$∴$ $\frac{x - y}{(1 + x y) x} + \frac{(x^{2} + 1) z^{2}}{x^{2} (z^{2} + 1)} = 1$

$∴$ $x (x - y) (z^{2} + 1) + z^{2} (1 + x^{2}) (1 + x y) = (1 + x y) x^{2} (1 + z^{2})$

$after solving we get$

$z^{2} = x y$ $∵$ $1 + x^{2} \neq 0$

$∴$ $\tan^{2} C = \tan A \cdot \tan B$

$∴$ $\tan A, \tan C, \tan B are in G.P.$

Q 21 :

If $\frac{3 π}{4} < α < π,$ then $\sqrt{\cos e c^{2} α + 2 c o t α}$ is equal to

$1 + c o t α$
$1 - c o t α$
$- 1 - c o t α$
$- 1 + c o t α$

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(3)

We have, $\sqrt{c o s e c^{2} α + 2 \cot α}$

$= \sqrt{1 + \cot^{2} α + 2 \cot α}$

$=$ $| 1 + \cot α |$

But $\frac{3 π}{4} < α < π$

$\Rightarrow$ $\cot α < - 1$

$\Rightarrow$ $1 + \cot α < 0$

Hence, $| 1 + \cot α |$ $= - (1 + \cot α)$

Q 22 :

The least value of $\cos e c^{2} x + 25 s e c^{2} x$ is

0
26
28
36

1

2

3

4

(4)

$c o s e c^{2} x + 25 \sec^{2} x$

$= 1 + \cot^{2} x + 25 (1 + \tan^{2} x)$

$= 26 + \cot^{2} x + 25 \tan^{2} x$

$= 26 + 10 + {(\cot x - 5 \tan x)}^{2} \geq 36$

Q 23 :

$(1 + \cos \frac{π}{8}) (1 + \cos \frac{3 π}{8}) (1 + \cos \frac{5 π}{8}) (1 + \cos \frac{7 π}{8})$ is equal to

1
$\cos \frac{π}{8}$
$\frac{1}{8}$
$\frac{1 + \sqrt{2}}{2 \sqrt{2}}$

1

2

3

4

(3)

Given expression,

$= (1 + \cos \frac{π}{8}) (1 + \cos \frac{3 π}{8}) (1 - \cos \frac{3 π}{8}) (1 - \cos \frac{π}{8})$

$= (1 - \cos^{2} \frac{π}{8}) (1 - \cos^{2} \frac{3 π}{8})$

$= \sin^{2} \frac{π}{8} \sin^{2} \frac{3 π}{8} = \sin^{2} \frac{π}{8} \cos^{2} \frac{π}{8}$

$= \frac{1}{4} \sin^{2} \frac{π}{4} = \frac{1}{8}$

Q 24 :

If $\sin θ = - \frac{4}{5}$ and $θ$ lies in the third quadrant, then $\cos \frac{θ}{2}$ is equal to

$\frac{1}{\sqrt{5}}$
$- \frac{1}{\sqrt{5}}$
$\sqrt{\frac{2}{5}}$
$- \sqrt{\frac{2}{5}}$

1

2

3

4

(2)

Given that, $\sin θ = - \frac{4}{5}$ and $θ$ lies in the 3rd quadrant.

$∴$ $\cos θ = - \sqrt{1 - \frac{16}{25}} = - \frac{3}{5}$

Now, $\cos \frac{θ}{2} = \pm \sqrt{\frac{1 + \cos θ}{2}}$

$= \pm \sqrt{\frac{1 - \frac{3}{5}}{2}} = \pm \sqrt{\frac{1}{5}}$

But we take $\cos \frac{θ}{2} = - \frac{1}{\sqrt{5}} .$

Since, if $θ$ lies in the 3rd quadrant, then $\frac{θ}{2}$ will be in the 2nd quadrant.

Hence, $\cos \frac{θ}{2} = - \frac{1}{\sqrt{5}}$

Q 25 :

If $\sin (α + β) = 1$ and $\sin (α - β) = \frac{1}{2},$ then $\tan (α + 2 β) \tan (2 α + β)$ is equal to

1
- 1
0
None of these

1

2

3

4

(1)

Since, $\sin (α + β) = 1$

$∴$ $α + β = \frac{π}{2}$ $\dots (i)$

and $\sin (α - β) = \frac{1}{2}$

$\Rightarrow$ $α - β = \frac{π}{6}$ $\dots (i i)$

On solving Eqs. (i) and (ii), we get

$α = \frac{π}{3}$ and $β = \frac{π}{6}$

Now, $\tan (α + 2 β) \tan (2 α + β)$

$= \tan (\frac{2 π}{3})$ $\tan (\frac{5 π}{6})$

$= \tan (π - \frac{π}{3})$ $\tan (π - \frac{π}{6})$

$= (- \cot \frac{π}{3})$ $(- \cot \frac{π}{6})$

$= \frac{1}{\sqrt{3}} \times \sqrt{3} = 1$

Q 26 :

If $0 < A < B < π,$ $\sin A + \sin B = \sqrt{\frac{3}{2}}$ and $\cos A + \cos B = \frac{1}{\sqrt{2}},$ then $A$ is equal to

$15 °$
$30 °$
$45 °$
$22 \frac{1}{2} °$

1

2

3

4

(1)

Clearly, ${(\sin A + \sin B)}^{2} + {(\cos A + \cos B)}^{2} = 2$

$∴$ $2 + 2 (\sin A \sin B + \cos A \cos B) = 2$

$\Rightarrow \cos (B - A) = 0 \Rightarrow B = A + 90 °$

Second equation gives,

$\cos A - \sin A = \frac{1}{\sqrt{2}}$

$\Rightarrow \cos (A + 45 °) = \cos 60 °$

$∴$ $A = 15 °$

Q 27 :

In a $∆ P Q R,$ $∠ R = \frac{π}{2} .$ If $\tan (\frac{P}{2})$ and $\tan (\frac{Q}{2})$ are the roots of $a x^{2} + b x + c = 0,$ where $a \neq 0,$ then

$b = a + c$
$b = c$
$c = a + b$
$a = b + c$

1

2

3

4

(3)

Since, $\tan \frac{P}{2} + \tan \frac{Q}{2} = \frac{- b}{a}$

and $\tan \frac{P}{2} \tan \frac{Q}{2} = \frac{c}{a}$

Also, $\frac{P}{2} + \frac{Q}{2} + \frac{R}{2} = \frac{π}{2}$

$\Rightarrow \frac{P + Q}{2} = \frac{π}{4}$ $[∵ ∠ R = \frac{π}{2}]$

$\Rightarrow \tan (\frac{P}{2} + \frac{Q}{2}) = 1$

$\Rightarrow \frac{\tan \frac{P}{2} + \tan \frac{Q}{2}}{1 - \tan \frac{P}{2} \tan \frac{Q}{2}} = 1$

$\Rightarrow \frac{- \frac{b}{a}}{1 - \frac{c}{a}} = 1$

$\Rightarrow$ $c = a + b$

Q 28 :

If $\cos (α + β) = \frac{4}{5}$ and $\sin (α - β) = \frac{5}{13},$ where $0 \leq α, β \leq \frac{π}{4},$ then $\tan 2 α$ is equal to

$\frac{25}{16}$
$\frac{56}{33}$
$\frac{19}{12}$
$\frac{20}{7}$

1

2

3

4

(2)

$\cos (α + β) = \frac{4}{5}$

$\Rightarrow$ $α + β \in 1st quadrant$

$\sin (α - β) = \frac{5}{13}$

$\Rightarrow$ $(α - β) \in 1st quadrant$

Now, $2 α = (α + β) + (α - β)$

$∴$ $\tan 2 α = \frac{\tan (α + β) + \tan (α - β)}{1 - \tan (α + β) \tan (α - β)}$

$= \frac{\frac{3}{4} + \frac{5}{12}}{1 - \frac{3}{4} \cdot \frac{5}{12}} = \frac{\frac{56}{48}}{\frac{33}{48}} = \frac{56}{33}$

Q 29 :

If $\tan α = {(1 + 2^{- x})}^{- 1}$ and $\tan β = {(1 + 2^{x + 1})}^{- 1},$ then $α + β$ equals

$\frac{π}{6}$
$\frac{π}{4}$
$\frac{π}{3}$
$\frac{π}{2}$

1

2

3

4

(2)

We know,

$\tan (α + β) = \frac{\tan α + \tan β}{1 - \tan α \tan β}$

$∵$ $\tan α = \frac{1}{1 + 2^{- x}}$

and $\tan β = \frac{1}{1 + 2^{x + 1}}$

$∴$ $\tan$ $(α + β)$ $= \frac{\frac{1}{1 + \frac{1}{2^{x}}} + \frac{1}{1 + 2^{x + 1}}}{1 - \frac{1}{1 + \frac{1}{2^{x}}} \cdot \frac{1}{1 + 2^{x + 1}}}$

$\Rightarrow \tan (α + β) = \frac{2^{x} + 2 \cdot 2^{2 x} + 2^{x} + 1}{1 + 2^{x} + 2 \cdot 2^{x} + 2 \cdot 2^{2 x} - 2^{x}}$

$\Rightarrow \tan (α + β) = 1$

$\Rightarrow$ $α + β = \frac{π}{4}$

Q 30 :

$cosec 10 ° - \sqrt{3} \sec 10 °$ is equal to

$1$
$\frac{1}{2}$
$2$
$4$

1

2

3

4

(4)

$cosec 10 ° - \sqrt{3} \sec 10 °$

$= \frac{1}{\sin 10 °} - \frac{\sqrt{3}}{\cos 10 °}$

$= (\frac{\cos 10 °}{2} - \frac{\sqrt{3}}{2} \sin 10 °) \frac{4}{\sin 20 °}$

$= \frac{4 \sin (30 ° - 10 °)}{\sin 20 °} = 4$

Topic Question Set

Q 1 :

If $\sin x = - \frac{3}{5},$ where $π < x < \frac{3 π}{2},$ then $80 (\tan^{2} x - \cos x)$ is equal to [2024]

Q 2 :

If the value of $\frac{3 \cos 36^{o} + 5 \sin 18^{o}}{5 \cos 36^{o} - 3 \sin 18^{o}}$ is $\frac{a \sqrt{5} - b}{c},$ where $a, b, c$ are natural numbers and $g c d (a, c) = 1,$ then $a + b + c$ is equal to : [2024]

Q 3 :

If $\tan A = \frac{1}{\sqrt{x (x^{2} + x + 1)}},$ $\tan B = \frac{\sqrt{x}}{\sqrt{x^{2} + x + 1}}$ and $\tan C = {(x^{- 3} + x^{- 2} + x^{- 1})}^{1 / 2}, 0 < A, B, C < π / 2,$ then $A + B$ is equal to: [2024]

Q 4 :

Let $A = {θ \in [0, 2 π] : 1 + 10 Re (\frac{2 \cos θ + i \sin θ}{\cos θ - 3 i \sin θ}) = 0}$ .

Then $\sum_{θ \in A} θ^{2}$ is equal to [2025]

Q 5 :

The value of (sin 70°) (cot 10° cot 70° – 1) is [2025]

Q 6 :

If $\sum_{r = 1}^{13} {\frac{1}{\sin (\frac{π}{4} + (r - 1) \frac{π}{6}) \sin (\frac{π}{4} + \frac{r π}{6})}} = a \sqrt{3} + b, b \in Z,$ then $a^{2} + b^{2}$ is equal to : [2025]

Q 8 :

If $\sin x + \sin^{2} x = 1, x \in (0, \frac{π}{2})$ , then $(\cos^{12} x + \tan^{12} x) + 3 (\cos^{10} x + \tan^{10} x + \cos^{8} x + \tan^{8} x) + (\cos^{6} x + \tan^{6} x)$ is equal to : [2025]

Q 9 :

The value of $36 (4 \cos^{2} 9 ° - 1) (4 \cos^{2} 27 ° - 1) (4 \cos^{2} 81 ° - 1) (4 \cos^{2} 243 ° - 1)$ is [2023]

Q 10 :

$96 \cos \frac{π}{33} \cos \frac{2 π}{33} \cos \frac{4 π}{33} \cos \frac{8 π}{33} \cos \frac{16 π}{33}$ is equal to [2023]

Q 11 :

The value of $\tan 9 ° - \tan 27 ° - \tan 63 ° + \tan 81 °$ is ________ . [2023]

Q 12 :

Let $α$ and $β$ respectively be the maximum and the minimum values of the function

$f (θ) = 4 (\sin^{4} (\frac{7 π}{2} - θ) + \sin^{4} (11 π + θ)) - 2 (\sin^{6} (\frac{3 π}{2} - θ) + \sin^{6} (9 π - θ)), θ \in R .$

Then $α + 2 β$ is equal to: [2026]

Q 13 :

The value of cosec $10^{o} - \sqrt{3}$ sec 10° is equal to [2026]

Q 14 :

If $\cot x = \frac{5}{12}$ for some $x \in (π, \frac{3 π}{2}),$ then $\sin 7 x (\cos \frac{13 x}{2} + \sin \frac{13 x}{2}) + \cos 7 x (\cos \frac{13 x}{2} - \sin \frac{13 x}{2})$ is equal to [2026]

Q 15 :

The value of $\frac{\sqrt{3} c o s e c 20 ° - \sec 20 °}{\cos 20 ° \cos 40 ° \cos 60 ° \cos 80 °}$ is equal to: [2026]

Q 16 :

If $\frac{\cos^{2} 48 ° - \sin^{2} 12 °}{\sin^{2} 24 ° - \sin^{2} 6 °} = \frac{α + β \sqrt{5}}{2},$ where $α, β \in ℕ$ , then $α + β$ is equal to __________ . [2026]

Q 17 :

The number of elements in the set ${x \in [0, 180 °] : \tan (x + 100 °) = \tan (x + 50 °) \tan x \tan (x - 50 °)}$ is __________ . [2026]

Q 18 :

Let $\frac{π}{2} < θ < π$ and $\cot θ = - \frac{1}{2 \sqrt{2}}$ . Then the value of $\sin (\frac{15 θ}{2}) (\cos 8 θ + \sin 8 θ) + \cos (\frac{15 θ}{2}) (\cos 8 θ - \sin 8 θ)$ is equal to [2026]

Q 19 :

Let $\cos (α + β) = - \frac{1}{10}$ and $\sin (α - β) = \frac{3}{8}$ , where $0 < α < \frac{π}{3}$ and $0 < β < \frac{π}{4}$ , If $\tan 2 α = \frac{3 (1 - r \sqrt{5})}{\sqrt{11} (s + \sqrt{5})}, r, s \in ℕ,$ then r + s is equal to______. [2026]

Q 20 :

If $\frac{\tan (A - B)}{\tan A} + \frac{\sin^{2} C}{\sin^{2} A} = 1, A, B, C \in ((0, \frac{π}{2}))$ , then [2026]

Q 21 :

If $\frac{3 π}{4} < α < π,$ then $\sqrt{\cos e c^{2} α + 2 c o t α}$ is equal to

Q 22 :

The least value of $\cos e c^{2} x + 25 s e c^{2} x$ is

Q 23 :

$(1 + \cos \frac{π}{8}) (1 + \cos \frac{3 π}{8}) (1 + \cos \frac{5 π}{8}) (1 + \cos \frac{7 π}{8})$ is equal to

Q 24 :

If $\sin θ = - \frac{4}{5}$ and $θ$ lies in the third quadrant, then $\cos \frac{θ}{2}$ is equal to

Q 25 :

If $\sin (α + β) = 1$ and $\sin (α - β) = \frac{1}{2},$ then $\tan (α + 2 β) \tan (2 α + β)$ is equal to

Q 26 :

If $0 < A < B < π,$ $\sin A + \sin B = \sqrt{\frac{3}{2}}$ and $\cos A + \cos B = \frac{1}{\sqrt{2}},$ then $A$ is equal to

Q 27 :

In a $∆ P Q R,$ $∠ R = \frac{π}{2} .$ If $\tan (\frac{P}{2})$ and $\tan (\frac{Q}{2})$ are the roots of $a x^{2} + b x + c = 0,$ where $a \neq 0,$ then

Q 28 :

If $\cos (α + β) = \frac{4}{5}$ and $\sin (α - β) = \frac{5}{13},$ where $0 \leq α, β \leq \frac{π}{4},$ then $\tan 2 α$ is equal to

Q 29 :

If $\tan α = {(1 + 2^{- x})}^{- 1}$ and $\tan β = {(1 + 2^{x + 1})}^{- 1},$ then $α + β$ equals

Q 30 :

$cosec 10 ° - \sqrt{3} \sec 10 °$ is equal to

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Topic Question Set

Q 1 : If sinx=-35, where π<x<3π2, then 80(tan2x-cosx) is equal to [2024]

Q 2 : If the value of 3cos36o+5sin18o5cos36o-3sin18o is a5-bc, where a,b,c are natural numbers and gcd(a,c)=1, then a+b+c is equal to : [2024]

Q 3 : If tanA=1x(x2+x+1), tanB=xx2+x+1 and tanC=(x-3+x-2+x-1)1/2,0<A,B,C<π/2, then A+B is equal to: [2024]

Q 4 : Let A={θ∈[0,2π]:1+10Re(2 cos θ+i sin θcos θ–3i sin θ)=0}. Then ∑θ∈Aθ2 is equal to [2025]

Q 5 : The value of (sin 70°) (cot 10° cot 70° – 1) is [2025]

Q 6 : If ∑r=113{1sin(π4+(r–1)π6)sin(π4+rπ6)}=a3+b, b∈Z, then a2+b2 is equal to : [2025]

Q 8 : If sinx+sin2x=1,x∈(0,π2), then (cos12x+tan12x)+3(cos10x+tan10x+cos8x+tan8x)+(cos6x+tan6x) is equal to : [2025]

Q 9 : The value of 36 (4cos29°-1)(4cos227°-1)(4cos281°-1)(4cos2243°-1) is [2023]

Q 10 : 96cosπ33cos2π33cos4π33cos8π33cos16π33 is equal to [2023]

Q 11 : The value of tan9°-tan27°-tan63°+tan81° is ________ . [2023]

Q 12 : Let α and β respectively be the maximum and the minimum values of the function f(θ)=4(sin4(7π2-θ)+sin4(11π+θ))-2(sin6(3π2-θ)+sin6(9π-θ)), θ∈R. Then α+2β is equal to: [2026]

Q 13 : The value of cosec 10o−3 sec 10° is equal to [2026]

Q 14 : If cotx=512 for some x∈(π,3π2), then sin7x(cos13x2+sin13x2)+cos7x(cos13x2-sin13x2) is equal to [2026]

Q 15 : The value of 3 cosec20°-sec20°cos20°cos40°cos60°cos80° is equal to: [2026]

Q 16 : If cos248°-sin212°sin224°-sin26°=α+β52, where α,β∈ℕ, then α+β is equal to __________ . [2026]

Q 17 : The number of elements in the set {x∈[0,180°]:tan(x+100°)=tan(x+50°)tan x tan(x-50°)} is __________ . [2026]

Q 18 : Let π2<θ<π and cotθ=-122. Then the value of sin(15θ2)(cos 8θ+sin 8θ)+cos(15θ2)(cos 8θ-sin 8θ) is equal to [2026]

Q 19 : Let cos(α+β)=-110 and sin(α-β)=38, where 0<α<π3 and 0<β<π4, If tan2α=3(1-r5)11(s+5), r,s∈ℕ, then r + s is equal to______. [2026]

Q 20 : If tan(A-B)tanA+sin2Csin2A=1, A,B,C∈(0,π2), then [2026]

Q 21 : If 3π4<α<π, then cosec2α+2 cot α is equal to

Q 22 : The least value of cosec2x+25 sec2 x is

Q 23 : (1+cosπ8)(1+cos3π8)(1+cos5π8)(1+cos7π8) is equal to

Q 24 : If sin θ=-45 and θ lies in the third quadrant, then cosθ2 is equal to

Q 25 : If sin(α+β)=1 and sin(α-β)=12, then tan(α+2β)tan(2α+β) is equal to

Q 26 : If 0<A<B<π, sinA+sinB=32 and cosA+cosB=12, then A is equal to

Q 27 : In a ∆PQR, ∠R=π2. If tan(P2) and tan(Q2) are the roots of ax2+bx+c=0, where a≠0, then

Q 28 : If cos(α+β)=45 and sin(α-β)=513, where 0≤α, β≤π4, then tan2α is equal to

Q 29 : If tanα=(1+2-x)-1 and tanβ=(1+2x+1)-1, then α+β equals

Q 30 : cosec10°-3sec10° is equal to

Want Us to Email you About Special Offers & Updates?

Q 1 :

If $\sin x = - \frac{3}{5},$ where $π < x < \frac{3 π}{2},$ then $80 (\tan^{2} x - \cos x)$ is equal to [2024]

Q 2 :

If the value of $\frac{3 \cos 36^{o} + 5 \sin 18^{o}}{5 \cos 36^{o} - 3 \sin 18^{o}}$ is $\frac{a \sqrt{5} - b}{c},$ where $a, b, c$ are natural numbers and $g c d (a, c) = 1,$ then $a + b + c$ is equal to : [2024]

Q 3 :

If $\tan A = \frac{1}{\sqrt{x (x^{2} + x + 1)}},$ $\tan B = \frac{\sqrt{x}}{\sqrt{x^{2} + x + 1}}$ and $\tan C = {(x^{- 3} + x^{- 2} + x^{- 1})}^{1 / 2}, 0 < A, B, C < π / 2,$ then $A + B$ is equal to: [2024]

Q 4 :

Let $A = {θ \in [0, 2 π] : 1 + 10 Re (\frac{2 \cos θ + i \sin θ}{\cos θ - 3 i \sin θ}) = 0}$ .

Then $\sum_{θ \in A} θ^{2}$ is equal to [2025]

Q 5 :

The value of (sin 70°) (cot 10° cot 70° – 1) is [2025]

Q 6 :

If $\sum_{r = 1}^{13} {\frac{1}{\sin (\frac{π}{4} + (r - 1) \frac{π}{6}) \sin (\frac{π}{4} + \frac{r π}{6})}} = a \sqrt{3} + b, b \in Z,$ then $a^{2} + b^{2}$ is equal to : [2025]

Q 8 :

If $\sin x + \sin^{2} x = 1, x \in (0, \frac{π}{2})$ , then $(\cos^{12} x + \tan^{12} x) + 3 (\cos^{10} x + \tan^{10} x + \cos^{8} x + \tan^{8} x) + (\cos^{6} x + \tan^{6} x)$ is equal to : [2025]

Q 9 :

The value of $36 (4 \cos^{2} 9 ° - 1) (4 \cos^{2} 27 ° - 1) (4 \cos^{2} 81 ° - 1) (4 \cos^{2} 243 ° - 1)$ is [2023]

Q 10 :

$96 \cos \frac{π}{33} \cos \frac{2 π}{33} \cos \frac{4 π}{33} \cos \frac{8 π}{33} \cos \frac{16 π}{33}$ is equal to [2023]

Q 11 :

The value of $\tan 9 ° - \tan 27 ° - \tan 63 ° + \tan 81 °$ is ________ . [2023]

Q 12 :

Let $α$ and $β$ respectively be the maximum and the minimum values of the function

$f (θ) = 4 (\sin^{4} (\frac{7 π}{2} - θ) + \sin^{4} (11 π + θ)) - 2 (\sin^{6} (\frac{3 π}{2} - θ) + \sin^{6} (9 π - θ)), θ \in R .$

Then $α + 2 β$ is equal to: [2026]

Q 13 :

The value of cosec $10^{o} - \sqrt{3}$ sec 10° is equal to [2026]

Q 14 :

If $\cot x = \frac{5}{12}$ for some $x \in (π, \frac{3 π}{2}),$ then $\sin 7 x (\cos \frac{13 x}{2} + \sin \frac{13 x}{2}) + \cos 7 x (\cos \frac{13 x}{2} - \sin \frac{13 x}{2})$ is equal to [2026]

Q 15 :

The value of $\frac{\sqrt{3} c o s e c 20 ° - \sec 20 °}{\cos 20 ° \cos 40 ° \cos 60 ° \cos 80 °}$ is equal to: [2026]

Q 16 :

If $\frac{\cos^{2} 48 ° - \sin^{2} 12 °}{\sin^{2} 24 ° - \sin^{2} 6 °} = \frac{α + β \sqrt{5}}{2},$ where $α, β \in ℕ$ , then $α + β$ is equal to __________ . [2026]

Q 17 :

The number of elements in the set ${x \in [0, 180 °] : \tan (x + 100 °) = \tan (x + 50 °) \tan x \tan (x - 50 °)}$ is __________ . [2026]

Q 18 :

Let $\frac{π}{2} < θ < π$ and $\cot θ = - \frac{1}{2 \sqrt{2}}$ . Then the value of $\sin (\frac{15 θ}{2}) (\cos 8 θ + \sin 8 θ) + \cos (\frac{15 θ}{2}) (\cos 8 θ - \sin 8 θ)$ is equal to [2026]

Q 19 :

Let $\cos (α + β) = - \frac{1}{10}$ and $\sin (α - β) = \frac{3}{8}$ , where $0 < α < \frac{π}{3}$ and $0 < β < \frac{π}{4}$ , If $\tan 2 α = \frac{3 (1 - r \sqrt{5})}{\sqrt{11} (s + \sqrt{5})}, r, s \in ℕ,$ then r + s is equal to______. [2026]

Q 20 :

If $\frac{\tan (A - B)}{\tan A} + \frac{\sin^{2} C}{\sin^{2} A} = 1, A, B, C \in ((0, \frac{π}{2}))$ , then [2026]

Q 21 :

If $\frac{3 π}{4} < α < π,$ then $\sqrt{\cos e c^{2} α + 2 c o t α}$ is equal to

Q 22 :

The least value of $\cos e c^{2} x + 25 s e c^{2} x$ is

Q 23 :

$(1 + \cos \frac{π}{8}) (1 + \cos \frac{3 π}{8}) (1 + \cos \frac{5 π}{8}) (1 + \cos \frac{7 π}{8})$ is equal to

Q 24 :

If $\sin θ = - \frac{4}{5}$ and $θ$ lies in the third quadrant, then $\cos \frac{θ}{2}$ is equal to

Q 25 :

If $\sin (α + β) = 1$ and $\sin (α - β) = \frac{1}{2},$ then $\tan (α + 2 β) \tan (2 α + β)$ is equal to

Q 26 :

If $0 < A < B < π,$ $\sin A + \sin B = \sqrt{\frac{3}{2}}$ and $\cos A + \cos B = \frac{1}{\sqrt{2}},$ then $A$ is equal to

Q 27 :

In a $∆ P Q R,$ $∠ R = \frac{π}{2} .$ If $\tan (\frac{P}{2})$ and $\tan (\frac{Q}{2})$ are the roots of $a x^{2} + b x + c = 0,$ where $a \neq 0,$ then

Q 28 :

If $\cos (α + β) = \frac{4}{5}$ and $\sin (α - β) = \frac{5}{13},$ where $0 \leq α, β \leq \frac{π}{4},$ then $\tan 2 α$ is equal to

Q 29 :

If $\tan α = {(1 + 2^{- x})}^{- 1}$ and $\tan β = {(1 + 2^{x + 1})}^{- 1},$ then $α + β$ equals

Q 30 :

$cosec 10 ° - \sqrt{3} \sec 10 °$ is equal to