MCQ
If ${\tan ^2}\theta = 2{\tan ^2}\phi + 1,$ then $\cos 2\theta + {\sin ^2}\phi $ equals
- A$-1$
- ✓$0$
- C$1$
- DNone of these
✓
Answer
Correct option: B.
$0$
b
(b) ${\tan ^2}\theta = 2{\tan ^2}\phi + 1 $
(b) ${\tan ^2}\theta = 2{\tan ^2}\phi + 1 $
$\Rightarrow 1 + {\tan ^2}\theta = 2\,(1 + {\tan ^2}\phi )$
==> ${\sec ^2}\theta = 2{\sec ^2}\phi $
$\Rightarrow {\cos ^2}\phi = 2{\cos ^2}\theta $
==> ${\cos ^2}\phi = 1 + \cos 2\theta $
$\Rightarrow {\sin ^2}\phi + \cos 2\theta = 0$.
Trick : Let $\theta = {45^o}$, then $\phi = 0$
$\therefore \;\cos (2 \times {45^o}) + {\sin ^2}0 = 0 + 0 = 0$.
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