Question

$\frac{1+\mathrm{cos\theta }-{\sin }^2\mathrm{\theta }}{\mathrm{sin\theta }(1+\mathrm{cos\theta })}=\mathrm{cot\theta }$

Answer 1

$\begin{array}{l}\\ \text{LHS=}\frac{1+\text{cos}\theta -{\text{sin}}^2\theta }{\text{sin}\theta (1+\text{cos}\theta )}\\ \text{=}\frac{(1+\text{cos}\theta )-(1-{\text{cos}}^2\theta )}{\text{sin}\theta (1+\text{cos}\theta )}\\ \text{=}\frac{\text{cos}\theta {\text{+cos}}^2\theta }{\text{sin}\theta (1+\text{cos}\theta )}\\ \text{=}\frac{\text{cos}\theta (1+\text{cos}\theta )}{\text{sin}\theta (1+\text{cos}\theta )}\\ \text{=}\frac{\text{cos}\theta }{\text{sin}\theta }\\ \text{=cot}\theta \\ \text{=RHS}\end{array}$
Hence, L.H.S. = R.H.S.