Question

If, $\tan 9^o=\frac{x}{y}$ then, value of $\frac{{\sec }^2{81}^o}{1+{\cot }^2{81}^o}$ is

• A. $\frac{x^3}{y^3}$
• B. $\frac{x^4}{y^4}$
• C. $\frac{x^5}{y^5}$
• D. $\frac{y^2}{x^2}$

Answer 1

Answer: (D)

$\frac{y^2}{x^2}$

Given,
$\tan 9^o=\frac{x}{y}$
$=\frac{{\sec }^2{81}^o}{1+{\cot }^2{81}^o}$
$=\frac{{\sec }^2{81}^o}{cose{c}^2{81}^o}$
$=\frac{\frac{1}{{\cos }^2{81}^o}}{\frac{1}{{\sin }^2{81}^o}}$
$=\frac{{\sin }^2{81}^o}{{\cos }^2{81}^o}$
$={\tan }^2{81}^o$
$=\left[\tan \right({90}^o-9^o){]}^2$
$=(\cot 9^o{)}^2$
$={\cot }^2{9}^o$
$=\frac{1}{{\tan }^2{9}^o}$
$=\frac{y^2}{x^2}$
Hence, option 'D' is correct.