Trigonometry – II — Maths STD 11 Science — Question

\begin{aligned}
& \text { Since, } 50^{\circ}=10^{\circ}+40^{\circ} \\
& \therefore \tan 50^{\circ}=\tan \left(10^{\circ}+40^{\circ}\right) \\
& \therefore \frac{\tan 10^{\circ}+\tan 40^{\circ}}{1-\tan 10^{\circ} \tan 40^{\circ}} \\
& \therefore \tan 50^{\circ}\left(1-\tan 10^{\circ} \tan 40^{\circ}\right)=\tan 10^{\circ}+\tan 40^{\circ} \\
& \therefore \tan 50^{\circ}-\tan 10^{\circ} \tan 40^{\circ} \tan 50^{\circ}=\tan 10^{\circ}+\tan 40^{\circ} \\
& \therefore \tan 50^{\circ}-\tan 10^{\circ} \tan 40^{\circ} \tan \left(90^{\circ}-40^{\circ}\right)=\tan 10^{\circ}+\tan \\
& 40^{\circ} \\
& \therefore \tan 50^{\circ}-\tan 10^{\circ} \tan 40^{\circ} \cot 40^{\circ} \\
& =\tan 10^{\circ}+\tan 40^{\circ} \ldots\left[\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right] \\
& \therefore \tan 50^{\circ}-\tan 10^{\circ} \tan 40^{\circ} \cdot \frac{1}{\tan 40^{\circ}}=\tan 10^{\circ}+\tan 40^{\circ} \\
& \therefore \tan 50^{\circ}-\tan 10^{\circ} \cdot 1=\tan 10^{\circ}+\tan 40^{\circ} \\
& \therefore \tan 50^{\circ}=\tan 40^{\circ}+2 \tan 10^{\circ}
\end{aligned}