Question

If $\sec \left(\theta \right)+\tan \left(\theta \right)=p,$ then find the value $\csc \left(\theta \right)$.

Answer 1

Given:

$\sec \left(\theta \right)+\tan \left(\theta \right)=p.....\left(1\right)$

Using Identity, $\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{2}}\left(\theta \right)\mathbf{-}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{2}}\left(\theta \right)\mathbf{=}\mathbf{1}$ ,

$\Rightarrow (\sec \left(\mathrm{\theta }\right)-\tan \left(\mathrm{\theta }\right))(\sec \left(\mathrm{\theta }\right)+\tan \left(\mathrm{\theta }\right))=1\left[\because a^2-b^2=(a+b)(a-b)\right]$

$$\begin{gathered} \Rightarrow \left(sec\right(\theta )-\tan (\theta \left)p\right)=1...(u\sin g(1\left)\right) \\ \Rightarrow \left(sec\right(\theta )-\tan (\theta \left)\right)=\frac{1}{p}...\left(2\right) \end{gathered}$$

Adding (1) and (2), we get,

$$\begin{gathered} \Rightarrow sec\left(\theta \right)+\tan \left(\theta \right)+sec\left(\theta \right)-\tan \left(\theta \right)=p+\frac{1}{p} \\ \Rightarrow 2\sec \left(\theta \right)=\frac{p^2+1}{p} \\ \Rightarrow \sec \left(\theta \right)=\frac{p^2+1}{2p} \\ \Rightarrow \cos \left(\theta \right)=\frac{2p}{p^2+1}\left[\because \cos \left(\theta \right)=\frac{1}{sec\left(\theta \right)}\right] \end{gathered}$$

Finding the value of $\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathit{\theta }\mathbf{\right)}$:

$$\begin{gathered} \sin \left(\theta \right)=\sqrt{1-{\cos }^2\left(\theta \right)}\mathbf{}\mathbf{}\mathbf{}\left[\because si{n}^2\left(\theta \right)+co{s}^2\left(\theta \right)=1\right] \\ \Rightarrow \sin \left(\theta \right)=\sqrt{1-{\left(\frac{2p}{p^2+1}\right)}^2} \\ \Rightarrow \sin \left(\theta \right)=\sqrt{\frac{{\left(p^2+1\right)}^2-{4p}^2}{{\left(p^2+1\right)}^2}} \\ \Rightarrow \sin \left(\theta \right)=\sqrt{\frac{{\left(p^2-1\right)}^2}{{\left(p^2+1\right)}^2}} \\ \Rightarrow \sin \left(\theta \right)=\frac{p^2-1}{p^2+1} \end{gathered}$$

Finding the value of $\mathit{c}\mathit{s}\mathit{c}\mathbf{\left(}\mathit{\theta }\mathbf{\right)}$:

$$\begin{gathered} \because \csc \left(\theta \right)=\frac{1}{\sin \left(\theta \right)} \\ \therefore \csc \left(\theta \right)=\frac{p^2+1}{p^2-1} \end{gathered}$$

Final answer: If $\sec \left(\theta \right)+\tan \left(\theta \right)=p,$then $\mathbf{csc}\left(\theta \right)\mathbf{=}\frac{{\mathbf{p}}^{\mathbf{2}}\mathbf{+}\mathbf{1}}{{\mathbf{p}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}$