Question

Prove:

$\left(i\right){\text{sin}}^2\theta +\frac{1}{1+{\text{tan}}^2\theta }=1$

$\left(ii\right)\frac{1}{1+{\text{tan}}^2\theta }+\frac{1}{1+{\text{cot}}^2\theta }=1$

Answer 1

$\left(i\right){\text{sin}}^2\theta +\frac{1}{1+{\text{tan}}^2\theta }=1$

LHS $={\text{sin}}^2\theta +\frac{1}{1+{\text{tan}}^2\theta }$

$={\text{sin}}^2\theta +\frac{1}{{\sec }^2\theta }$ $\because {\text{sec}}^2\theta -{\text{tan}}^2\theta =1$

$={\text{sin}}^2\theta +{\text{cos}}^2\theta$

$=1$ ($\because {\text{sin}}^2\theta +{\text{cos}}^2\theta =1$)

$=$RHS

(ii) $\frac{1}{1+{\text{tan}}^2\theta }+\frac{1}{1+{\text{cot}}^2\theta }=1$

LHS $=\frac{1}{1+{\text{tan}}^2\theta }+\frac{1}{1+{\text{cot}}^2\theta }$

$=\frac{1}{{\text{sec}}^2\theta }+\frac{1}{{\text{cosec}}^2\theta }$ ($\because {\text{sec}}^2\theta -{\text{tan}}^2\theta =1$ and ${\text{cosec}}^2\theta -{\text{cot}}^2\theta =1$

$={\text{cos}}^2\theta +{\text{sin}}^2\theta$

$=1$

$=$ RHS