Question

The value of$\frac{(3+\cot 76°.\cot 16°)}{(\cot 76°+\cot 16°)}$ is$\tan 46°$

Answer 1

Prove that the value of$\frac{(3+\cot 76°.\cot 16°)}{(\cot 76°+\cot 16°)}$ is$\tan 46°$

$\begin{array}{rcl}\frac{(3+\cot 76°.\cot 16°)}{(\cot 76°+\cot 16°)}& =& \frac{3\sin 76°\sin 16°+\cos 76°\cos 16°}{\cos 76°\sin 16°+\cos 16°\sin 76°}\\ & =& \frac{2\sin 76°\sin 16°+\sin 76°\sin 16°+\cos 76°\cos 16°}{\cos 76°\sin 16°+\cos 16°\sin 76°}\\ & =& \frac{2\sin 76°\sin 16°+\cos 60°}{\sin (76°+16°)}\\ & =& \frac{\cos 60°-\cos 92°+\cos 60°}{\sin 92°}\\ & =& \frac{1-\cos 92°}{\sin 92°}\\ & =& \frac{2{\sin }^{2}46°}{2\sin 46°\cos 46°}\\ & =& \tan 46°\\ \therefore \frac{(3+\cot 76°.\cot 16°)}{(\cot 76°+\cot 16°)}& =& \tan 46°\end{array}$

Hence it is proved.