Question

If cosec θ + cot θ = p, prove that cos θ = $\frac{(p^2-1)}{(p^2+1)}.$

Answer 1

$$\begin{gathered} \cos \text{ec}\theta +\cot \theta =p \\ =>\frac{1}{\sin \theta }+\frac{\cos \theta }{\sin \theta }=p \\ =>\frac{1+\cos \theta }{\sin \theta }=p \\ \text{Squaring both sides, we get:} \\ {\left(\frac{1+\cos \theta }{\sin \theta }\right)}^2=p^2 \\ =>\frac{{(1+\cos \theta )}^2}{{\sin }^2\theta }=p^2 \\ =>\frac{{(1+\cos \theta )}^2}{1-{\cos }^2\theta }=p^2 \\ =>\frac{{(1+\cos \theta )}^2}{(1+\cos \theta )(1-\cos \theta )}=p^2 \\ =>\frac{(1+\cos \theta )}{(1-\cos \theta )}=p^2 \\ =>1+\cos \theta =p^2(1-\cos \theta ) \\ =>1+\cos \theta =p^2-p^2\cos \theta \\ =>\cos \theta (1+p^2)=p^2-1 \\ =>\cos \theta =\frac{p^2-1}{p^2+1} \\ \text{Hence proved}\text{.} \end{gathered}$$