Question

Prove that: $\tan \left(\frac{\pi }{4}+\mathrm{\theta }\right)+\tan \left(\frac{\pi }{4}-\mathrm{\theta }\right)=2\sec 2\mathrm{\theta }$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\tan \left(\frac{\mathrm{\pi }}{4}+\mathrm{\theta }\right)+\tan \left(\frac{\mathrm{\pi }}{4}-\mathrm{\theta }\right) \\ =\frac{\tan {\displaystyle \frac{\mathrm{\pi }}{4}}+\mathrm{tan\theta }}{1-\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\mathrm{tan\theta }}+\frac{\tan {\displaystyle \frac{\mathrm{\pi }}{4}}-\mathrm{tan\theta }}{1+\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\mathrm{tan\theta }}\left[\because \tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}\mathrm{and}\tan \left(A-B\right)=\frac{\tan A-\tan B}{1+\tan A\tan B}\right] \end{gathered}$$

$$\begin{gathered} =\frac{1+\mathrm{tan\theta }}{1-\mathrm{tan\theta }}+\frac{1-\mathrm{tan\theta }}{1+\mathrm{tan\theta }} \\ =\frac{{\left(1+\mathrm{tan\theta }\right)}^2+{\left(1-\mathrm{tan\theta }\right)}^2}{\left(1+\mathrm{tan\theta }\right)\left(1-\mathrm{tan\theta }\right)} \\ =\frac{2(1+{\tan }^2\mathrm{\theta })}{\left(1-{\tan }^2\mathrm{\theta }\right)}=\frac{2\left({\sec }^2\mathrm{\theta }\right)}{1-{\displaystyle \frac{{\sin }^2\mathrm{\theta }}{{\cos }^2\mathrm{\theta }}}} \end{gathered}$$

$$\begin{gathered} =\frac{2\left({\sec }^2\mathrm{\theta }\right)\left({\cos }^2\mathrm{\theta }\right)}{\cos 2\mathrm{\theta }}\left(\because {\cos }^2\mathrm{\theta }-{\sin }^2\mathrm{\theta }=\cos 2\mathrm{\theta }\right) \\ =\frac{2\times 1}{\cos 2\mathrm{\theta }} \\ =2\sec 2\mathrm{\theta }=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$