Question

Solve:

$\frac{\left(cotA-1\right)}{\left(2-se{c}^{2}A\right)}=\frac{\left(cotA\right)}{\left(1+\tan A\right)}$

Answer 1

Proof of trigonometric identities:

$\frac{\left(cotA-1\right)}{\left(2-se{c}^{2}A\right)}=\frac{\left(cotA\right)}{\left(1+\tan A\right)}$

LHS:

$\frac{\left(cotA-1\right)}{\left(2-se{c}^{2}A\right)}=\frac{{\displaystyle \frac{1}{\tan A}}-1}{2-\left(1+{\tan }^{2}A\right)}$ $(\because {\sec }^{2}A=1+{\tan }^{2}A)$

$\frac{{\displaystyle \frac{1}{\tan A}}-1}{2-\left(1+{\tan }^{2}A\right)}=\frac{1-\tan A}{\tan A(1-{\tan }^{2}A)}$ $\left[\because \left(1-{\tan }^{2}A\right)=\left(1-\tan A\right)\left(1+\tan A\right)\right]$

$\frac{\left(1-\tan A\right)}{\tan A\left(1-\tan A\right)\left(1+\tan A\right)}=\frac{1}{\tan A\left(1+\tan A\right)}=\frac{cotA}{\left(1+\tan A\right)}$

LHS=RHS

Hence, it is proved that $\frac{\left(cotA-1\right)}{\left(2-se{c}^{2}A\right)}=\frac{\left(cotA\right)}{\left(1+\tan A\right)}$.