Question

If $\tan \left(\frac{\mathrm{\pi }}{4}+\mathrm{x}\right)+\tan \left(\frac{\mathrm{\pi }}{4}-\mathrm{x}\right)=\mathrm{a}$ then find the value of ${\tan }^2\left(\frac{\mathrm{\pi }}{4}+x\right)+{\tan }^2\left(\frac{\mathrm{\pi }}{4}-x\right)$.

Answer 1

Calculate the value of the given expression :

Given,

$\tan \left(\frac{\mathrm{\pi }}{4}+\mathrm{x}\right)+\tan \left(\frac{\mathrm{\pi }}{4}-\mathrm{x}\right)=\mathrm{a}$

$\begin{array}{cc}& {\left[\tan \left(\frac{\mathrm{\pi }}{4}+x\right)+\tan \left(\frac{\mathrm{\pi }}{4}+x\right)\right]}^2=a^2\\ \Rightarrow & {\tan }^2\left(\frac{\mathrm{\pi }}{4}+x\right)+{\tan }^2\left(\frac{\mathrm{\pi }}{4}-x\right)+2\tan \left(\frac{\mathrm{\pi }}{4}+x\right)\tan \left(\frac{\mathrm{\pi }}{4}-x\right)=a^2\left[\because {\left(a+b\right)}^2=a^2+b^2+2ab\right]...\left(\mathrm{i}\right)\end{array}$

Using $\tan \left(\mathrm{A}+\mathrm{B}\right)=\frac{\tan \left(\mathrm{A}\right)+\tan \left(\mathrm{B}\right)}{1-\tan \left(\mathrm{A}\right)\tan \left(\mathrm{B}\right)}$

$\begin{array}{ccc}\Rightarrow & \tan \left[\left(\frac{\mathrm{\pi }}{4}+x\right)+\left(\frac{\mathrm{\pi }}{4}-x\right)\right]=\frac{\tan \left(\frac{\mathrm{\pi }}{4}+x\right)+\tan \left(\frac{\mathrm{\pi }}{4}-x\right)}{1-\tan \left(\frac{\mathrm{\pi }}{4}+x\right)\tan \left(\frac{\mathrm{\pi }}{4}-x\right)}& \\ \Rightarrow & \tan \left(\frac{\mathrm{\pi }}{2}\right)=\frac{\tan \left({\displaystyle \frac{\mathrm{\pi }}{4}}+x\right)+\tan \left({\displaystyle \frac{\mathrm{\pi }}{4}}-x\right)}{1-\tan \left({\displaystyle \frac{\mathrm{\pi }}{4}}+x\right)\tan \left({\displaystyle \frac{\mathrm{\pi }}{4}}-x\right)}& \\ \Rightarrow & 1-\tan \left(\frac{\mathrm{\pi }}{4}+x\right)\tan \left(\frac{\mathrm{\pi }}{4}-x\right)=0& \left[\because \tan \left(\frac{\mathrm{\pi }}{2}\right)=\infty ,\infty =\frac{a}{0}\right]\\ \Rightarrow & \tan \left(\frac{\mathrm{\pi }}{4}+x\right)\tan \left(\frac{\mathrm{\pi }}{4}-x\right)=1& \end{array}$

Put value of $\tan \left(\frac{\mathrm{\pi }}{4}+x\right)\tan \left(\frac{\mathrm{\pi }}{4}-x\right)$ back in $\left(\mathrm{i}\right)$.

$\begin{array}{ccc}\Rightarrow & {\tan }^2\left(\frac{\mathrm{\pi }}{4}+x\right)+{\tan }^2\left(\frac{\mathrm{\pi }}{4}-x\right)+2=a^2& \left[\because \tan \left(\frac{\mathrm{\pi }}{4}+x\right)\tan \left(\frac{\mathrm{\pi }}{4}-x\right)=1\right]\\ \Rightarrow & {\tan }^2\left(\frac{\mathrm{\pi }}{4}+\mathrm{x}\right)+{\tan }^2\left(\frac{\mathrm{\pi }}{4}-\mathrm{x}\right)={\mathrm{a}}^2-2& \end{array}$

Hence, value of ${\tan }^2\left(\frac{\mathrm{\pi }}{4}+x\right)+{\tan }^2\left(\frac{\mathrm{\pi }}{4}-x\right)$ is ${\mathrm{a}}^2-2$.