Question

Solve

$\frac{2\sin 68}{\cos 22}-{\frac{2\cot 15}{5\tan {75}^{\circ }}}^{\circ }-\frac{3\tan {45}^{\circ }\tan {20}^{\circ }\tan 40\tan 50\tan {70}^{\circ }}{5}$

Answer 1

$\frac{2\sin {68}^{\circ }}{\cos {22}^{\circ }}-\frac{2\cot {15}^{\circ }}{5\tan {75}^{\circ }}-\frac{3\tan {45}^{\circ }\tan {20}^{\circ }\tan {40}^{\circ }\tan {50}^{\circ }\tan {70}^{\circ }}{5}$

$=\frac{2\sin {68}^{\circ }}{\cos ({90}^{\circ }-{68}^{\circ })}-\frac{2\cot {15}^{\circ }}{5\tan ({90}^{\circ }-{15}^{\circ })}-\frac{3\times 1(\tan {20}^{\circ }\cdot \tan {40}^{\circ })(\tan {50}^{\circ }\cdot \tan {70}^{\circ })}{5}$

$=\frac{2\sin {68}^{\circ }}{\sin {68}^{\circ }}-\frac{2\cot {15}^{\circ }}{5\cot {15}^{\circ }}-\frac{3}{5}(\tan {20}^{\circ }\cdot \tan {40}^{\circ })(\tan {50}^{\circ }\cdot \tan {70}^{\circ })$

$[\because \sin ({90}^{\circ }-\theta )=\cos \theta$ and $\tan ({90}^{\circ }-\theta )=\cot \theta ]$

$=2-\frac{2}{5}-\frac{3}{5}(\tan {20}^{\circ }\cdot \tan {40}^{\circ })(\tan {50}^{\circ }\cdot \tan {70}^{\circ })$

$=\frac{8}{5}-\frac{3}{5}(\tan {20}^{\circ }\cdot \tan {40}^{\circ }\cdot \tan ({90}^{\circ }-{40}^{\circ })\cdot \tan ({90}^{\circ }-{20}^{\circ }\left)\right)$

$=\frac{8}{5}-\frac{3}{5}(\tan {20}^{\circ }\cdot \tan {40}^{\circ }\cdot \cot {40}^{\circ }\cdot \cot {20}^{\circ })$

$=\frac{8}{5}-\frac{3}{5}(\tan {20}^{\circ }\cdot \cot {20}^{\circ }\cdot \tan {40}^{\circ }\cdot \cot {40}^{\circ })$

$=\frac{8}{5}-\frac{3}{5}$ $(\because \tan \theta \cdot \cot \theta =1)$

$=1$