Question

If $\alpha$ is the angle of first quadrant such that $co{\sec }^4\alpha =17+{\cot }^4\alpha$, then what is the value of $\sin \alpha$?

• A. $\frac{1}{3}$
• B. $\frac{1}{4}$
• C. $\frac{1}{9}$
• D. $\frac{1}{16}$

Answer 1

The correct option is A $\frac{1}{3}$

$cose{c}^4\alpha -co{t}^4\alpha =17$

(As $cose{c}^2\alpha -co{t}^2\alpha =1)$

$\Rightarrow (cose{c}^2\alpha -co{t}^2\alpha )(cose{c}^2\alpha +co{t}^2\alpha )=17$

$\Rightarrow cose{c}^2\alpha +co{t}^2\alpha =\mathrm{17...}\left(1\right)$

$cose{c}^2\alpha -co{t}^2\alpha =\mathrm{1...}\left(2\right)$

then $\left(1\right)+\left(2\right)\Rightarrow 2cose{c}^2\alpha =18$

$\Rightarrow si{n}^2\alpha =\frac{1}{9}\Rightarrow sin\alpha =\frac{1}{3}$ $(\because \alpha in1stquadrant)$