10(sin2θ)2+15(1−sin2θ)2=6 Let sin2θ=t⇒10t2+15(1−t)2=1610t2+15−30t+15t2=625t2−30t+9=0(5t−3)2=0sin2θ=53 and cos2θ=5216(25)427×27125+8+8125=125×5250=52
If 10sin4θ+15cos4θ=6, then the value of 16sec8θ27cosec6θ+8sec6θ is:
Held on 4 Apr 2025 · Verified 6 Jul 2026.
52
43
53
51
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
If $\cot x=\frac{5}{12}$ for some $x \in\left(\pi, \frac{3 \pi}{2}\right)$, then $\sin 7 x\left(\cos \frac{13 x}{2}+\sin \frac{13 x}{2}\right)+\cos 7 x\left(\cos \frac{13 x}{2}-\sin \frac{13 x}{2}\right)$ is equal to
The value of sin²30° + cos²30° is:
Considering the principal values of inverse trigonometric functions, the value of the expression $\tan \left(2 \sin ^{-1}\left(\frac{2}{\sqrt{13}}\right)-2 \cos ^{-1}\left(\frac{3}{\sqrt{10}}\right)\right)$ is equal to :
The number of solutions of 2sin²x + sin²2x = 2 in [0, 2π] is:
The value of $\frac{\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}}{\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}}$ is equal to
Work through every JEE Main Trigonometry PYQ, year by year.