Evaluate:
\( \left(\frac{3 \tan 41^{\circ}}{\cot 49^{\circ}}\right)^{2}-\left(\frac{\sin 35^{\circ} \sec 55^{\circ}}{\tan 10^{\circ} \tan 20^{\circ} \tan 60^{\circ} \tan 70^{\circ} \tan 80^{\circ}}\right)^{2} \)

Given:

\( \left(\frac{3 \tan 41^{\circ}}{\cot 49^{\circ}}\right)^{2}-\left(\frac{\sin 35^{\circ} \sec 55^{\circ}}{\tan 10^{\circ} \tan 20^{\circ} \tan 60^{\circ} \tan 70^{\circ} \tan 80^{\circ}}\right)^{2} \).

To do:

We have to evaluate \( \left(\frac{3 \tan 41^{\circ}}{\cot 49^{\circ}}\right)^{2}-\left(\frac{\sin 35^{\circ} \sec 55^{\circ}}{\tan 10^{\circ} \tan 20^{\circ} \tan 60^{\circ} \tan 70^{\circ} \tan 80^{\circ}}\right)^{2} \). 

Solution:  

We know that,

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

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

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

$cos\ \theta \times \sec\ \theta=1$

Therefore,

$\left(\frac{3 \tan 41^{\circ}}{\cot 49^{\circ}}\right)^{2}-\left(\frac{\sin 35^{\circ} \sec 55^{\circ}}{\tan 10^{\circ} \tan 20^{\circ} \tan 60^{\circ} \tan 70^{\circ} \tan 80^{\circ}}\right)^{2}=\left[\frac{3\tan( 90^{\circ}-49^{\circ})}{\cot 49^{\circ}}\right]^{2} -\left[\frac{\sin( 90^{\circ}-55^{\circ})\sec 55^{\circ}}{\tan 10^{\circ}\tan 20^{\circ}\left(\sqrt{3}\right)\tan( 90^{\circ}-20^{\circ})\tan( 90^{\circ}-10^{\circ})}\right]^{2}$

$=\left(\frac{\cot 49^{\circ}}{\cot 49^{\circ}}\right)^{2} -\left(\frac{\cos 55^{\circ}\sec 55^{\circ}}{\tan 10^{\circ}\tan 20^{\circ}\left(\sqrt{3}\right)\cot 20^{\circ}\cot 10^{\circ}}\right)^{2}$

$=( 1)^{2} -\left(\frac{1}{( 1)\left(\sqrt{3}\right)( 1)}\right)^{2}$

$=1-\left(\frac{1}{\sqrt{3}}\right)^{2}$

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

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

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

$=\frac{2}{3}$

Hence, $\left(\frac{3 \tan 41^{\circ}}{\cot 49^{\circ}}\right)^{2}-\left(\frac{\sin 35^{\circ} \sec 55^{\circ}}{\tan 10^{\circ} \tan 20^{\circ} \tan 60^{\circ} \tan 70^{\circ} \tan 80^{\circ}}\right)^{2}=\frac{2}{3}$. 

Tutorialspoint

Updated on: 10-Oct-2022

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