Question

$1\1 + \tan( \alpha {?}^{2} )$
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Answer 1

Answer:

we know that

$1 + \tan ^{2} \alpha = \sec ^{2} \alpha$

and

$\frac{1}{ \sec^{2} \alpha } = \cos^{2} \alpha$

so,

$\frac{1}{1 + {tan}^{2} \alpha }$

$= \frac{1}{sec ^{2} \alpha }$

$= \cos ^{2} ( \alpha )$