Using cos2A−sin2B=cos(A+B)cos(A−B):
cos248°−sin212°=cos60°cos36°=21cos36°.
Using sin2A−sin2B=sin(A+B)sin(A−B):
sin224°−sin26°=sin30°sin18°=21sin18°.
Ratio =sin18°cos36°=(5−1)/4(1+5)/4=4(1+5)2=46+25=23+5.
α=3, β=1. α+β=4.
JEE Main 2026 — Mathematics Trigonometry
If sin224∘−sin26∘cos248∘−sin212∘=2α+β5, where α,β∈N, then α+β is equal to _____
Held on 22 Jan 2026 · Verified 6 Jul 2026.
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
The number of elements in the set $\left\{x \in\left[0,180^{\circ}\right]: \tan \left(x+100^{\circ}\right)=\tan \left(x+50^{\circ}\right) \tan x \tan \left(x-50^{\circ}\right)\right\}$ is $\_\_\_\_$.
The value of sin²30° + cos²30° is:
Let $\alpha$ and $\beta$ respectively be the maximum and the minimum values of the function $f(\theta)=4\left(\sin ^{4}\left(\frac{7 \pi}{2}-\theta\right)+\sin ^{4}(11 \pi+\theta)\right)-2\left(\sin ^{6}\left(\frac{3 \pi}{2}-\theta\right)+\sin ^{6}(9 \pi-\theta)\right), \theta \in \mathbf{R}$. Then $\alpha+2 \beta$ is equal to :
Number of solutions of $\sqrt{3} \cos 2 \theta+8 \cos \theta+3 \sqrt{3}=0, \theta \in[-3 \pi, 2 \pi]$ is:
If $\frac{\tan (\mathrm{A}-\mathrm{B})}{\tan \mathrm{A}}+\frac{\sin ^{2} \mathrm{C}}{\sin ^{2} \mathrm{~A}}=1, \mathrm{~A}, \mathrm{~B}, \mathrm{C} \in\left(0, \frac{\pi}{2}\right)$, then
Work through every JEE Main Trigonometry PYQ, year by year.