Question

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

Answer 1

We know,

$\cos (90-\theta )=\sin \theta$

$\tan (90-\theta )=\cot \theta$

$\cot \theta =\frac{1}{\tan \theta }$

$\cos {22}^{\circ }=\cos ({90}^{\circ }-{68}^{\circ })=\sin {68}^{\circ }$

$\tan {75}^{\circ }=\tan ({90}^{\circ }-{15}^{\circ })=\cot {15}^{\circ }$

$\tan {20}^{\circ }=\tan ({90}^{\circ }-{70}^{\circ })=\cot {70}^{\circ }$

$\tan {40}^{\circ }=\tan ({90}^{\circ }-{50}^{\circ })=\cot {50}^{\circ }$

To evaluate,

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

$=\frac{2\sin {68}^{\circ }}{\sin {68}^{\circ }}-\frac{2\cot {15}^{\circ }}{5\cot {15}^{\circ }}-\frac{3\tan {45}^{\circ }\cot {70}^{\circ }\cot {50}^{\circ }\tan {50}^{\circ }\tan {70}^{\circ }}{5}$

$=\frac{2\sin {68}^{\circ }}{\sin {68}^{\circ }}-\frac{2\cot {15}^{\circ }}{5\cot {15}^{\circ }}-\frac{3\tan {45}^{\circ }\tan {50}^{\circ }\tan {70}^{\circ }}{5\tan {70}^{\circ }\tan {50}^{\circ }}$

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

$=2-\frac{(2+3)}{5}$

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

$=2-1$

$=1$