Question

If $\mathrm{cosec}\left(\theta \right)+\cot \left(\theta \right)=p$ then prove that $\cos \left(\theta \right)=\frac{p^2-1}{p^2+1}$

Answer 1

Prove the given expression

Given,

$\mathrm{cosec}\left(\theta \right)+\cot \left(\theta \right)=p..........\left(1\right)$

Now,

$$\begin{gathered} {\mathrm{cosec}}^2\left(\theta \right)-{\cot }^2\left(\theta \right)=1 \\ \Rightarrow \left(\mathrm{cosec}\left(\theta \right)-\cot \left(\theta \right)\right)\left(\mathrm{cosec}\left(\theta \right)+\cot \left(\theta \right)\right)=1 \\ \Rightarrow \left(\mathrm{cosec}\left(\theta \right)-\cot \left(\theta \right)\right)p=1 \\ \Rightarrow \left(\mathrm{cosec}\left(\theta \right)-\cot \left(\theta \right)\right)=\frac{1}{p}...........\left(2\right) \end{gathered}$$

Adding $\left(1\right)$ and $\left(2\right)$ we get

$$\begin{gathered} 2\mathrm{cosec}\left(\theta \right)=p+\frac{1}{p} \\ \Rightarrow \mathrm{cosec}\left(\theta \right)=\frac{p^2+1}{2p} \\ \Rightarrow \sin \left(\theta \right)=\frac{2p}{p^2+1}\left[\because \sin \left(A\right)=\frac{1}{\mathrm{cosec}\left(A\right)}\right] \\ \Rightarrow \cos \left(\theta \right)=\sqrt{1-{\left(\frac{2p}{p^2+1}\right)}^2}\left[\because \cos \left(A\right)=\sqrt{1-{\sin }^2\left(A\right)}\right] \\ \Rightarrow \cos \left(\theta \right)=\sqrt{\frac{p^4+2p^2+1-4p^2}{{\left(p^2+1\right)}^2}} \\ \Rightarrow \cos \left(\theta \right)=\sqrt{\frac{p^4-2p^2+1}{{\left(p^2+1\right)}^2}} \\ \Rightarrow \cos \left(\theta \right)=\sqrt{\frac{{\left(p^2-1\right)}^2}{{\left(p^2+1\right)}^2}} \\ \Rightarrow \cos \left(\theta \right)=\frac{\left(p^2-1\right)}{\left(p^2+1\right)} \end{gathered}$$

Hence proved, $\cos \left(\theta \right)=\frac{p^2-1}{p^2+1}$.