Question

$\frac{{\cot }^2\mathrm{\theta }(\mathrm{sec\theta }-1)}{(1+\mathrm{sin\theta })}={\sec }^2\mathrm{\theta }.\left(\frac{1-\mathrm{sin\theta }}{1+\mathrm{sec\theta }}\right)$

Answer 1

$\begin{array}{l}\\ \text{LHS=}\frac{{\text{cot}}^2\theta (\text{sec}\theta -1)}{(1+\text{sin}\theta )}\\ \text{=}\frac{\frac{{\text{cos}}^2\theta }{{\text{sin}}^2\theta }(\frac{1}{\text{cos}\theta }-1)}{(1+\text{sin}\theta )}\\ \text{=}\frac{\frac{\text{cos}\theta }{{\text{sin}}^2\theta }-\frac{{\text{cos}}^2\theta }{{\text{sin}}^2\theta }}{(1+\text{sin}\theta )}\\ \text{=}\frac{\text{cos}\theta (1-\text{cos}\theta )}{{\text{sin}}^2\theta (1+\text{sin}\theta )}\\ \text{=}\frac{\text{cos}\theta (1-\text{cos}\theta )}{(1-{\text{cos}}^2\theta )(1+\text{sin}\theta )}\\ \text{=}\frac{\text{cos}\theta (1-\text{cos}\theta )}{\left(\text{1+cos}\theta \right)(1-\text{cos}\theta )(1+\text{sin}\theta )}\\ \text{=}\frac{\text{cos}\theta }{(1+\frac{1}{\text{sec}\theta })(1+\text{sin}\theta )}\\ \text{=}\frac{\text{cos}\theta \text{sec}\theta }{\left(\text{sec}\theta \text{+1}\right)(1+\text{sin}\theta )}\\ \text{=}\frac{1}{\left(\text{sec}\theta \text{+1}\right)(1+\text{sin}\theta )}\\ \text{=}\frac{(1-\text{sin}\theta )}{\left(\text{sec}\theta \text{+1}\right)(1+\text{sin}\theta )(1-\text{sin}\theta )}\text{[Multiplying the numerator and denominator by (1-sinθ)]}\\ \text{=}\frac{(1-\text{sin}\theta )}{\left(\text{sec}\theta \text{+1}\right)(1-{\sin }^2\theta )}\\ \text{=}\frac{(1-\text{sin}\theta )}{\left(\text{sec}\theta \text{+1}\right){\text{cos}}^2\theta }\\ \text{=}\frac{{\text{sec}}^2\theta (1-\text{sin}\theta )}{\left(\text{sec}\theta \text{+1}\right)}\\ \text{=RHS}\end{array}$
Hence, LHS = RHS