Trigonometric Ratios and Identities — JEE Main PYQs

68 previous year questions on Trigonometric Ratios and Identities (Mathematics), newest first. Open any question for its complete step-by-step solution.

The value of $cosec 1 0^{\circ} - 3 sec 1 0^{\circ}$ is equal to:
JEE Main 2026View solution →
If $\frac{c o s ^{2} 4 8 ^{\circ} - s i n ^{2} 1 2 ^{\circ}}{s i n ^{2} 2 4 ^{\circ} - s i n ^{2} 6 ^{\circ}} = \frac{α + β 5}{2}$ , where $α, β \in N$ , then $α + β$ is equal to
JEE Main 2026View solution →
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 $tan 2 α = \frac{3 ( 1 - r 5 )}{11 ( s + 5 )}, r, s \in N$ , then $r + s$ is equal to ______。
JEE Main 2026View solution →
Let $α$ and $β$ respectively be the maximum and the minimum values of the function …
JEE Main 2026View solution →
Let $\frac{π}{2} < θ < π$ and $cot θ = - \frac{1}{2 2}$ . 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 :
JEE Main 2026View solution →
The least value of $(cos^{2} θ - 6 sin θ cos θ + 3 sin^{2} θ + 2)$ is
JEE Main 2026View solution →
The value of $\frac{3 cosec 2 0 ^{\circ} - s e c 2 0 ^{\circ}}{c o s 2 0 ^{\circ} c o s 4 0 ^{\circ} c o s 6 0 ^{\circ} c o s 8 0 ^{\circ}}$ is equal to
JEE Main 2026View solution →
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
JEE Main 2026View solution →
If $sin (\frac{π}{18}) sin (\frac{5 π}{18}) sin (\frac{7 π}{18}) = K$ , then the value of $sin (\frac{10 K π}{3})$ is :
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If $A = \frac{s i n 3 ^{\circ}}{c o s 9 ^{\circ}} + \frac{s i n 9 ^{\circ}}{c o s 2 7 ^{\circ}} + \frac{s i n 2 7 ^{\circ}}{c o s 8 1 ^{\circ}}$ and $B = tan 8 1^{\circ} - tan 3^{\circ}$ , then $\frac{B}{A}$ is equal to ____.
JEE Main 2026View solution →
Let $tan A, tan B$ , where $A, B \in (- \frac{π}{2}, \frac{π}{2})$ , be the roots of the quadratic equation $x^{2} - 2 x - 5 = 0$ . Then $20 sin^{2} (\frac{A + B}{2})$ is equal to :
JEE Main 2026View solution →
The sum of all the integral values of $p$ such that the equation $3 sin^{2} x + 12 cos x - 3 = p, x \in R$ , has at least one solution, is :
JEE Main 2026View solution →
Let $P = {θ \in [0, 4 π] : tan^{2} θ \neq = 1}$ and $S = {a \in Z : 2 (cos^{8} θ - sin^{8} θ) sec 2 θ = a^{2}, θ \in P}$ . Then $n (S)$ is :
JEE Main 2026View solution →
The value of $(sin 7 0^{\circ}) (cot 1 0^{\circ} cot 7 0^{\circ} - 1)$ is
JEE Main 2025View solution →
Let the range of the function
$f (x) = 6 + 16 cos x \cdot cos (\frac{π}{3} - x) \cdot cos (\frac{π}{3} + x) \cdot sin 3 x \cdot cos 6 x, x \in R$ be $[α, β]$ . Then the distance of the point $(α, β)$ from the line …
JEE Main 2025View solution →
If $sin x + sin^{2} x = 1, x \in (0, \frac{π}{2})$ , then …
JEE Main 2025View solution →
If $10 sin^{4} θ + 15 cos^{4} θ = 6$ , then the value of $\frac{27 cosec ^{6} θ + 8 s e c ^{6} θ}{16 s e c ^{8} θ}$ is:
JEE Main 2025View solution →
If for $θ \in [- \frac{π}{3}, 0]$ , the points
$(x, y) = (3 tan (θ + \frac{π}{3}), 2 tan (θ + \frac{π}{6}))$ lie on
$xy + α x + β y + γ = 0$ , then $α^{2} + β^{2} + γ^{2}$ …
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If $sin x = - \frac{3}{5}$ , where $π < x < \frac{3 π}{2}$ ,
then $80 (tan^{2} x - cos x)$ is equal to :
JEE Main 2024View solution →
If the range of $f (θ) = \frac{s i n ^{4} θ + 3 c o s ^{2} θ}{s i n ^{4} θ + c o s ^{2} θ}, θ \in R$ is $[α, β]$ , then the sum of the infinite G.P., whose first term is 64 and the common ratio is $\frac{α}{β}$ , is equal to _____ …
JEE Main 2024View solution →
If the value of $\frac{3 c o s 3 6 ^{\circ} + 5 s i n 1 8 ^{\circ}}{5 c o s 3 6 ^{\circ} - 3 s i n 1 8 ^{\circ}}$ is $\frac{a 5 - b}{c}$ , where $a, b, c$ are natural numbers and $gcd (a, c) = 1$ , then …
JEE Main 2024View solution →
Let
$∣ cos θ cos (60 - θ) cos (60 + θ) ∣ \leq \frac{1}{8}, θ \in [0, 2 π]$
Then, the sum of all $θ \in [0, 2 π]$ , where $cos 3 θ$ attains its maximum value, is :
JEE Main 2024View solution →
Let the range of the function
$f (x) = \frac{1}{2 + s i n 3 x + c o s 3 x}, x \in IR$ be $[a, b]$ .
If $α$ and $β$ are respectively the A.M. and the G.M. of $a$ and $b$ , then $\frac{α}{β}$ is equal to :
JEE Main 2024View solution →
The sum of the solutions $x \in R$ of the equation $\frac{3 c o s 2 x + c o s ^{3} 2 x}{c o s ^{6} x - s i n ^{6} x} = x^{3} - x^{2} + 6$ is
JEE Main 2024View solution →
If $tan A = \frac{1}{x ( x ^{2} + x + 1 )}, tan B = \frac{x}{x ^{2} + x + 1}$ and $tan C = (x^{- 3} + x^{- 2} + x^{- 1})^{\frac{1}{2}}, 0 < A, B, C < \frac{π}{2}$ , then $A + B$ is equal to :
JEE Main 2024View solution →
The value of $tan 9^{\circ} - tan 2 7^{\circ} - tan 6 3^{\circ} + tan 8 1^{\circ}$ is
JEE Main 2023View solution →
The value of $36 (4 cos^{2} 9^{\circ} - 1) (4 cos^{2} 2 7^{\circ} - 1) (4 cos^{2} 8 1^{\circ} - 1) (4 cos^{2} 24 3^{\circ} - 1)$ is
JEE Main 2023View solution →
$96 cos \frac{π}{33} cos \frac{2 π}{33} cos \frac{4 π}{33} cos \frac{8 π}{33} cos \frac{16 π}{33}$ is equal to
JEE Main 2023View solution →
If $tan 1 5^{\circ} + \frac{1}{t a n 7 5 ^{\circ}} + \frac{1}{t a n 10 5 ^{\circ}} + tan 19 5^{\circ} = 2 a$ , then the value of $(a + \frac{1}{a})$ is:
JEE Main 2023View solution →
The set of all values of $λ$ for which the equation $cos^{2} 2 x - 2 sin^{4} x - 2 cos^{2} x = λ$
JEE Main 2023View solution →
Let $f (θ) = 3 (sin^{4} (\frac{3 π}{2} - θ) + sin^{4} (3 π + θ)) - 2 (1 - sin^{2} 2 θ)$ and
$S = {θ \in [0, π] : f^{'} (θ) = - \frac{3}{2}}$ . If $4 β = \sum_{θ \in S} θ$ , then …
JEE Main 2023View solution →
Let $S = {θ \in (0, \frac{π}{2}) : \sum_{m = 1}^{9} sec (θ + (m - 1) \frac{π}{6}) sec (θ + \frac{mπ}{6}) = - \frac{8}{3}}$

Then
JEE Main 2022View solution →
$2 sin (\frac{π}{22}) sin (\frac{3 π}{22}) sin (\frac{5 π}{22}) sin (\frac{7 π}{22}) sin (\frac{9 π}{22})$ is equal to
JEE Main 2022View solution →
If $cot α = 1$ and $sec β = - \frac{5}{3}$ , where $π < α < \frac{3 π}{2}$ and $\frac{π}{2} < β < π$ , then the value of $tan (α + β)$ and the quadrant in which $α + β$ lies, respectively are
JEE Main 2022View solution →
$α = sin 3 6^{\circ}$ is a root of which of the following equation
JEE Main 2022View solution →
$16 sin (2 0^{\circ}) sin (4 0^{\circ}) sin (8 0^{\circ})$ is equal to :
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If $sin^{2} (1 0^{\circ}) sin (2 0^{\circ}) sin (4 0^{\circ}) sin (5 0^{\circ}) sin (7 0^{\circ}) = α - \frac{1}{16} sin (1 0^{\circ})$ , then $16 + α^{- 1}$ is equal to ____.
JEE Main 2022View solution →
The value of $cos (\frac{2 π}{7}) + cos (\frac{4 π}{7}) + cos (\frac{6 π}{7})$ is equal to :
JEE Main 2022View solution →
The value of $2 sin (1 2^{\circ}) - sin (7 2^{\circ})$ is :
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The value of $2 sin (\frac{π}{8}) sin (\frac{2 π}{8}) sin (\frac{3 π}{8}) sin (\frac{5 π}{8}) sin (\frac{6 π}{8}) sin (\frac{7 π}{8})$ is :
JEE Main 2021View solution →
The number of integral values of ' $k$ ' for which the equation $3 sin x + 4 cos x = k + 1$ has a solution, $k \in R$ is ______
JEE Main 2021View solution →
$cosec 1 8^{\circ}$
is a root of the equation:
JEE Main 2021View solution →
If $sin θ + cos θ = \frac{1}{2}$ , then $16 (sin (2 θ) + cos (4 θ) + sin (6 θ))$ is equal to :
JEE Main 2021View solution →
The value of $cot \frac{π}{24}$ is :
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If for $x \in (0, \frac{π}{2}), lo g_{10} sin x + lo g_{10} cos x = - 1$ and $lo g_{10} (sin x + cos x) = \frac{1}{2} (lo g_{10}^{n} - 1), n > 0$ , then the value of $n$ is equal to :
JEE Main 2021View solution →
If $15 sin^{4} α + 10 cos^{4} α = 6$ , for some $α \in R$ , then the value of $27 sec^{6} α + 8 cosec^{6} α$ is equal to :
JEE Main 2021View solution →
Let $tan α, tan β$ and $tan γ; α, β, γ \neq = \frac{( 2 n - 1 ) π}{2}, n \in N$ be the slopes of three line segments $O A, O B$ and OC , respectively, where O is origin. If circumcentre of $△ ABC$ coincides with origin and its …
JEE Main 2021View solution →
If $\frac{2 s i n α}{1 + c o s 2 α} = \frac{1}{7}$ and $\frac{1 - c o s 2 β}{2} = \frac{1}{10}, α, β \in (0, \frac{π}{2})$ , then $tan (α + 2 β)$ is equal to ______.
JEE Main 2020View solution →
The value of $cos^{3} (\frac{π}{8}) \cdot cos (\frac{3 π}{8}) + sin^{3} (\frac{π}{8}) \cdot sin (\frac{3 π}{8})$ is:
JEE Main 2020View solution →
If $L = sin^{2} (\frac{π}{16}) - sin^{2} (\frac{π}{8})$ and $M = cos^{2} (\frac{π}{16}) - sin^{2} (\frac{π}{8})$ , then:
JEE Main 2020View solution →