Question

Prove the following trigonometric identities:

$si{n}^{2}A+\frac{1}{1+ta{n}^{2}A}=1$

Answer 1

We know that,

$co{s}^{2}A+si{n}^{2}A=1$

$se{c}^{2}A-ta{n}^{2}A=1$

So, $si{n}^{2}A+\frac{1}{1+ta{n}^{2}A}=si{n}^{2}A+\frac{1}{se{c}^{2}A}$

=$si{n}^{2}A+\frac{1}{(secA{)}^2}$ = $si{n}^2\theta +co{s}^2\theta$

= 1