Question

If $\tan ^{-1}\left(\sin ^2 \theta+2 \sin \theta+2\right)+\cot ^{-1}\left(4^{\sec ^2 \phi}+1\right)=\frac{\pi}{2}$ has solution for some $\theta$ and $\phi$ then

• A. $\sin \theta=-1$
• B. $\sin \theta=1$
• C. $ \cos \phi=-1$
• D. $\cos \phi=1$

Answer 1

Answer: (B), (C), (D)

For given equation to hold true
$\sin ^2 \theta+2 \sin \theta+2=4^{\sec ^2 \phi+1} $
$\text { L.H.S. }=(\sin \theta+1)^2+1 $
$\therefore 1 \leq(\sin \theta+1)^2+1 \leq 5 \text { and RHS } \geq 5 \text { as } \sec ^2 \phi \geq 1 $
$\therefore \text { LHS }=\text { RHS }=5 $
$\therefore \sin \theta=1 \text { and } \sec ^2 \phi=1 \Rightarrow \cos \phi= \pm 1 $