Question

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

Answer 1

$\Rightarrow$$\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}$

$=\frac{2\sin ({90}^{\circ }-{22}^{\circ })}{\cos {22}^{\circ }}-\frac{2\cot ({90}^{\circ }-{75}^{\circ })}{5\tan {75}^{\circ }}-\frac{3\left(1\right)\left(\tan {20}^{\circ }\tan {70}^{\circ }\right)\left(\tan {40}^{\circ }\tan {50}^{\circ }\right)}{5}$

$=\frac{2\cos {22}^{\circ }}{\cos {22}^{\circ }}-\frac{2\tan {75}^{\circ }}{5\tan {75}^{\circ }}-\frac{3\left[\tan \right({90}^{\circ }-{70}^{\circ }\left)\tan {70}^{\circ }\right]\left[\tan \right({90}^{\circ }-{50}^{\circ }\left)\tan {50}^{\circ }\right]}{5}$

$=2-\frac{2}{3}-\frac{3}{5}\left(\cot {70}^{\circ }\tan {70}^{\circ }\right)\left(\cot {50}^{\circ }\tan {50}^{\circ }\right)$

$[\therefore \tan ({90}^{\circ }-\theta )=\cot \theta ,\cot ({90}^{\circ }-\theta )=\tan \theta \&\sin ({90}^{\circ }-\theta )=\cos \theta ]$

$=2-\frac{2}{5}-\frac{3}{5}=2-1=1$