Question

If $sec\left(A\right)+\tan \left(A\right)=p$ then find the value of $\cos ec\left(A\right)$.

Answer 1

Step $1:$ Compute the values of $sec\left(A\right),\tan \left(A\right)$.

Since it is given that $sec\left(A\right)+\tan \left(A\right)=p...1$.

Multiply and divide the left-hand side by $sec\left(A\right)-\tan \left(A\right)$.

$$\begin{gathered} \Rightarrow \left(sec\left(A\right)+\tan \left(A\right)\right)\times \frac{sec\left(A\right)-\tan \left(A\right)}{sec\left(A\right)-\tan \left(A\right)}=p \\ \Rightarrow \frac{se{c}^2\left(A\right)-{\tan }^2\left(A\right)}{sec\left(A\right)-\tan \left(A\right)}=p \\ \Rightarrow \frac{1}{sec\left(A\right)-\tan \left(A\right)}=p \\ \Rightarrow sec\left(A\right)-\tan \left(A\right)=\frac{1}{p}...2 \end{gathered}$$ {Since, $se{c}^2\left(\theta \right)-{\tan }^2\left(\theta \right)=1$}

Add equation $1$ and equation $2$.

$$\begin{gathered} sec\left(A\right)+\tan \left(A\right)+sec\left(A\right)-\tan \left(A\right)=p+\frac{1}{p} \\ \Rightarrow 2sec\left(A\right)=\frac{p^2+1}{p} \end{gathered}$$

Therefore, $sec\left(A\right)=\frac{p^2+1}{2p}$.

Subtract equation $1$ and equation $2$.

$$\begin{gathered} sec\left(A\right)+\tan \left(A\right)-sec\left(A\right)+\tan \left(A\right)=p-\frac{1}{p} \\ \Rightarrow 2\tan \left(A\right)=\frac{p^2-1}{p} \end{gathered}$$

Therefore, $\tan \left(A\right)=\frac{p^2-1}{2p}$.

Step $2:$ Find the required value.

As we know that, $\cos ec\left(\theta \right)=\frac{1}{\sin \left(\theta \right)}$.

So, $\cos ec\left(A\right)=\frac{1}{\sin \left(A\right)}$

Multiply and divide the right-hand side by $\cos \left(A\right)$

.$$\begin{gathered} \Rightarrow \cos ec\left(A\right)=\frac{1}{\sin \left(A\right)}\times \frac{\cos \left(A\right)}{\cos \left(A\right)} \\ \Rightarrow \cos ec\left(A\right)=\frac{sec\left(A\right)}{\tan \left(A\right)} \end{gathered}$$ {Since, $\frac{\cos \left(\theta \right)}{\sin \left(\theta \right)}=\frac{1}{\tan \left(\theta \right)}$ and $\frac{1}{\cos \left(\theta \right)}=sec\left(\theta \right)$}

Substitute the value of $sec\left(A\right)$ and $\tan \left(A\right)$.

$$\begin{gathered} \Rightarrow \cos ec\left(A\right)=\frac{p^2+1}{2p}\times \frac{2p}{p^2-1} \\ \Rightarrow \cos ec\left(A\right)=\frac{p^2+1}{p^2-1} \end{gathered}$$

Hence, the value of $\cos ec\left(A\right)$ is $\frac{p^2+1}{p^2-1}$.