1 + sec 2 x sin 2 x =

1 + \sec^{2} x \sin^{2} x =

$\sin 2 x$
$\sin^{2} x$
$\tan^{2} x$
$\sec^{2} x$

Solution

Let
$f (x) = 1 + \sec^{2} x \sin^{2} x$
$\Rightarrow f (x) = 1 + (\frac{\sin^{2} x}{\cos^{2} x})$
$[∵ \sec x = \frac{1}{\cos x}]$
$\Rightarrow f (x) = 1 + \tan^{2} x$
$[∵ \tan x = \frac{\sin x}{\cos x}]$
$\Rightarrow f (x) = \sec^{2} x$

Asked in: AP EAMCET 2022 (04 Jul Shift 2)

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