Question

If $\tan \left(\pi /4+\mathrm{\theta }\right)+\tan \left(\pi /4-\mathrm{\theta }\right)=\lambda \sec 2\mathrm{\theta },\mathrm{then}\lambda =$
(a) 3
(b) 4
(c) 1
(d) 2

Answer 1

(d) 2

$$\begin{gathered} \mathrm{Given}: \\ \tan \left(\frac{\mathrm{\pi }}{4}+\mathrm{\theta }\right)+\tan \left(\frac{\mathrm{\pi }}{4}-\mathrm{\theta }\right)=\mathrm{\lambda }\sec 2\mathrm{\theta } \\ \Rightarrow \frac{\tan {\displaystyle \frac{\mathrm{\pi }}{4}}+\mathrm{tan\theta }}{1-\tan \frac{\mathrm{\pi }}{4}\times \mathrm{tan\theta }}+\frac{\tan {\displaystyle \frac{\mathrm{\pi }}{4}}-\mathrm{tan\theta }}{1+\tan \frac{\mathrm{\pi }}{4}\times \mathrm{tan\theta }}=\mathrm{\lambda }\sec 2\mathrm{\theta } \\ \Rightarrow \frac{1+\mathrm{tan\theta }}{1-\mathrm{tan\theta }}+\frac{1-\mathrm{tan\theta }}{1+\mathrm{tan\theta }}=\mathrm{\lambda }\sec 2\mathrm{\theta } \\ \Rightarrow \frac{{\left(1+\mathrm{tan\theta }\right)}^2+{\left(1-\mathrm{tan\theta }\right)}^2}{\left(1-\mathrm{tan\theta }\right)\left(1+\mathrm{tan\theta }\right)}=\mathrm{\lambda }\sec 2\mathrm{\theta } \\ \Rightarrow \frac{2\left(1+{\tan }^2\mathrm{\theta }\right)}{1-{\tan }^2\mathrm{\theta }}=\mathrm{\lambda }\sec 2\mathrm{\theta } \end{gathered}$$

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