Question

Question 9
If $sinA+si{n}^{2}A=1$, then the value of $(co{s}^{2}A+co{s}^{4}A)$ is

(A) $1$
(B) $\frac{1}{2}$
(C) $2$
(D) $3$

Answer 1

The answer is (A).

Given, $sinA+si{n}^{2}A=1$

$\Rightarrow sinA=1-si{n}^{2}A=co{s}^{2}A[\because si{n}^{\theta }+co{s}^2\theta =1]$

$Onsquaringbothsidesweget,$

$si{n}^{2}A=co{s}^{4}A$

$\Rightarrow 1-co{s}^{2}A=co{s}^{4}A$

$\Rightarrow co{s}^{2}A+co{s}^{4}A=1(Ans.)$.