Question : If $ \frac{k-k \cot ^2 30^{\circ}}{1+\cot ^2 30^{\circ}}=\sin ^2 60^{\circ}+4 \tan ^2 45^{\circ}-\operatorname{cosec}^2 60^{\circ}$, then the value of k (correct to two decimal places) is:
Option 1: 5.55
Option 2: – 6.83
Option 3: – 5.58
Option 4: 6.83
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Correct Answer: – 6.83
Solution :
Computing RHS,
$\sin ^2 60^{\circ}+4 \tan ^2 45^{\circ}-\operatorname{cosec}^2 60^{\circ} = (\frac{\sqrt{3}}{2})^2+4\times1^2-(\frac{2}{\sqrt{3}})^2$
⇒ $\sin ^2 60^{\circ}+4 \tan ^2 45^{\circ}-\operatorname{cosec}^2 60^{\circ} = \frac{3}{4}+4-\frac{4}{3}$
⇒ $\sin ^2 60^{\circ}+4 \tan ^2 45^{\circ}-\operatorname{cosec}^2 60^{\circ} = \frac{41}{12}$
Computing LHS,
$ \frac{k-k \cot ^2 30^{\circ}}{1+\cot ^2 30^{\circ}} = k(\frac{1- (\sqrt{3})^2}{1+(\sqrt{3})^2})$
⇒ $ \frac{k-k \cot ^2 30^{\circ}}{1+\cot ^2 30^{\circ}} = \frac{-k}{2}$
So, $\frac{-k}{2}=\frac{41}{12}$
⇒ $k=-6.83$
Hence, the correct answer is – 6.83.
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