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} $

Given:

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

To do:

We have to evaluate \( \frac{2 \sin 68^{\circ}}{\cos 22^{\circ}}-\frac{2 \cot 15^{\circ}}{5 \tan 75^{\circ}}-\frac{3 \tan 45^{\circ} \tan 20^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 70^{\circ}}{5} \). 

Solution:  

We know that,

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

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

$tan\ \theta \times \cot\ \theta=1$

Therefore,

$\frac{2 \sin 68^{\circ}}{\cos 22^{\circ}}-\frac{2 \cot 15^{\circ}}{5 \tan 75^{\circ}}-\frac{3 \tan 45^{\circ} \tan 20^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 70^{\circ}}{5}=\frac{2\sin( 90^{\circ}-22^{\circ})}{\cos 22^{\circ}} -\frac{2\cot 15^{\circ}}{5\tan( 90^{\circ}-15^{\circ})} -\frac{3( 1)\tan 20^{\circ}\tan 40^{\circ}\tan( 90^{\circ}-40^{\circ})\tan( 90^{\circ}-20^{\circ})}{5}$

$=\frac{2\cos 22^{\circ}}{\cos 22^{\circ}} -\frac{2\cot 15^{\circ}}{5\cot 15^{\circ}} -\frac{3\tan 20^{\circ}\tan 40^{\circ}\cot 50^{\circ}\cot 20^{\circ}}{5}$

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

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

$=\frac{10-5}{5}$

$=\frac{5}{5}$

$=1$

Hence, $\frac{2 \sin 68^{\circ}}{\cos 22^{\circ}}-\frac{2 \cot 15^{\circ}}{5 \tan 75^{\circ}}-\frac{3 \tan 45^{\circ} \tan 20^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 70^{\circ}}{5}=1$.

Tutorialspoint

Updated on: 10-Oct-2022

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