# Solve the following

Question:

Evaluate:

$\frac{\sec \theta \operatorname{cosec}\left(90^{\circ}-\theta\right)-\tan \theta \cot \left(90^{\circ}-\theta\right)+\sin ^{2} 55^{\circ}+\sin ^{2} 35^{\circ}}{\tan 10^{\circ} \tan 20^{*} \tan 60^{\circ} \tan 70^{\circ} \tan 80^{\circ}}$

Solution:

Here we to evaluate the value of the expression given as follow

$\frac{\sec \theta \operatorname{cosec}\left(90^{\circ}-\theta\right)-\tan \theta \cot \left(90^{\circ}-\theta\right)+\sin ^{2} 55^{\circ}+\sin ^{2} 35^{\circ}}{\tan 10^{\circ} \tan 20^{\circ} \tan 60^{\circ} \tan 70^{\circ} \tan 80^{\circ}}$

Now we are using the following identities

$\operatorname{cosec}\left(90^{\circ}-\theta\right)=\sec \theta$

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

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

Therefore the given expression can be written as

$\frac{\sec \theta \sec \theta-\tan \theta \tan \theta+\sin ^{2} 55^{\circ}+\cos ^{2}\left(90^{\circ}-55^{\circ}\right)}{\tan 10^{\circ} \tan 20^{\circ} \tan 60^{\circ} \cot \left(90^{\circ}-20^{\circ}\right) \cot \left(90^{\circ}-10^{\circ}\right)}$

$=\frac{\sec ^{2} \theta-\tan ^{2} \theta+\left(\sin ^{2} 55^{\circ}+\cos ^{2} 55^{\circ}\right)}{\left(\tan 10^{\circ} \cot 10^{\circ}\right)\left(\tan 20^{\circ} \cot 20^{\circ}\right) \times \sqrt{3}}$$=\frac{1+1}{1 \times 1 \times \sqrt{3}}$

$=\frac{1+1}{1 \times 1 \times \sqrt{3}}$

$=\frac{2}{\sqrt{3}}$

Hence the value of the given expression is $\frac{2}{\sqrt{3}}$