Question

If $\tan 9^o=\frac{q}{p}$ then prove that $\frac{{\sec }^2{81}^o}{1+{\cot }^2{81}^o}=\frac{p^2}{q^2}$.

Answer 1

Given,

$\tan 9^o=\frac{q}{p}$.....(1).

Now,

$\frac{{\sec }^2{81}^o}{1+{\cot }^2{81}^o}$

$=\frac{1+{\tan }^2{81}^o}{1+{\cot }^2{81}^o}$ [ Since ${\sec }^2\theta +{\tan }^2\theta =1$]

$=\frac{1+{\cot }^2{9}^o}{1+{\tan }^2{9}^o}$ [ Since $\tan {81}^o=\cot 9^o$] and $\cot {81}^o=\tan 9^o$]

$=\frac{1+\frac{p^2}{q^2}}{1+\frac{q^2}{p^2}}$ [ Using (1)]

$=\frac{p^2}{q^2}$.