Question

If $\tan 9^o=\frac{P}{q}$, then the value of $\frac{{\sec }^2{81}^o}{1+{\cot }^2{81}^o}$ ?

Answer 1

$\tan A=\tan 9^0=\frac{BC}{BA}=\frac{p}{q}$

$A{C}^2=B{C}^2+B{A}^2=AC=\sqrt{p^2+q^2}$

$\sec 81=\frac{CA}{CB}=\frac{\sqrt{p^2+q^2}}{p}$

$\cot 81=\frac{p}{q}$

$\frac{{\sec }^{2}81}{1+{\cot }^{2}81}=\frac{p^2+q^2}{1+\frac{p^2}{q^2}}=\frac{q^2(p^2+q^2)}{p^2+q^2}=q^2$