Evaluate each of the following:$ \frac{\sin 30^{\circ}-\sin 90^{\circ}+2 \cos 0^{\circ}}{\tan 30^{\circ} \tan 60^{\circ}} $

Given:

\( \frac{\sin 30^{\circ}-\sin 90^{\circ}+2 \cos 0^{\circ}}{\tan 30^{\circ} \tan 60^{\circ}} \)

To do:

We have to evaluate \( \frac{\sin 30^{\circ}-\sin 90^{\circ}+2 \cos 0^{\circ}}{\tan 30^{\circ} \tan 60^{\circ}} \).

Solution:  

We know that,

$sin 30^{\circ}=\frac{1}{2}$

$\sin 90^{\circ}=1$

$\cos 0^{\circ}=1$

$\tan 30^{\circ}=\frac{1}{\sqrt3}$

$\tan 60^{\circ}=\sqrt3$

$\frac{\sin 30^{\circ}-\sin 90^{\circ}+2 \cos 0^{\circ}}{\tan 30^{\circ} \tan 60^{\circ}}=\frac{\left(\frac{1}{2}\right) -1+2( 1)}{\left(\frac{1}{\sqrt{3}}\right)\left(\sqrt{3}\right)}$

$=\frac{\frac{1}{2} +1}{1}$

$=\frac{1+2( 1)}{2}$

$=\frac{3}{2}$

Hence, $\frac{\sin 30^{\circ}-\sin 90^{\circ}+2 \cos 0^{\circ}}{\tan 30^{\circ} \tan 60^{\circ}}=\frac{3}{2}$. 

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Updated on: 10-Oct-2022

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