Let A={θ:sinθ=tanθ} and B={θ:cosθ=1} Now, A={θ:sinθ=cosθsinθ} ={θ:sinθ(cosθ−1)=0} ={θ=0,π,2π,3π,……} For B:cosθ=1⇒θ=π,2π,4π,…… This shows that A is not contained in B. i.e. A⊂B. but B⊂A.
Let A=θ: sin (θ)=tan (θ) and B=(θ: cos (θ)= 1\} be two sets. Then:
JEE Main 2013 — Mathematics Trigonometry
Let A={θ:sin(θ)=tan(θ)} and B=(θ:cos(θ)= 1} be two sets. Then:
Held on 25 Apr 2013 · Verified 6 Jul 2026.
Options
- A
A=B
- B
A⊂B
- C
B⊂A
- D
A⊂B and B−A=ϕ
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
JEE Main Trigonometry in other years
More JEE Main Trigonometry questions
Number of solutions of $\sqrt{3} \cos 2 \theta+8 \cos \theta+3 \sqrt{3}=0, \theta \in[-3 \pi, 2 \pi]$ is:
The value of sin²30° + cos²30° is:
If $\frac{\cos ^{2} 48^{\circ}-\sin ^{2} 12^{\circ}}{\sin ^{2} 24^{\circ}-\sin ^{2} 6^{\circ}}=\frac{\alpha+\beta \sqrt{5}}{2}$, where $\alpha, \beta \in \mathbb{N}$, then $\alpha+\beta$ is equal to $\_\_\_\_\_$
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 :
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 $\_\_\_\_$.
Other Mathematics topics for JEE Main
Keep practicing Trigonometry
Work through every JEE Main Trigonometry PYQ, year by year.