Evaluate: cos left(sec ^{-1} 2right)+tan left | vedclass

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Question

(C) We know that $\sec^{-1}(2) = \frac{\pi}{3}$ because $\sec(\frac{\pi}{3}) = 2$. We know that $\cot^{-1}(\sqrt{3}) = \frac{\pi}{6}$ because $\cot(\f

Vedclass · Accepted Answer

$\frac{5+\sqrt{3}}{2 \sqrt{3}}$

Vedclass · Answer

(C) We know that $\sec^{-1}(2) = \frac{\pi}{3}$ because $\sec(\frac{\pi}{3}) = 2$. We know that $\cot^{-1}(\sqrt{3}) = \frac{\pi}{6}$ because $\cot(\frac{\pi}{6}) = \sqrt{3}$. We know that $\operatorname{cosec}^{-1}(\frac{2}{\sqrt{3}}) = \frac{\pi}{3}$ because $\operatorname{cosec}(\frac{\pi}{3}) = \frac{2}{\sqrt{3}}$. Substituting these values into the expression: $\cos(\frac{\pi}{3}) + 	an(\frac{\pi}{6}) + \sin(\frac{\pi}{3}) = \frac{1}{2} + \frac{1}{\sqrt{3}} + \frac{\sqrt{3}}{2}$. Combining the terms: $\frac{1}{2} + \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{3}} = \frac{1+\sqrt{3}}{2} + \frac{1}{\sqrt{3}} = \frac{\sqrt{3}(1+\sqrt{3}) + 2}{2\sqrt{3}} = \frac{\sqrt{3} + 3 + 2}{2\sqrt{3}} = \frac{5+\sqrt{3}}{2\sqrt{3}}$. Thus,the correct option is $C$.

Evaluate: $\cos \left(\sec ^{-1} 2\right)+\tan \left(\cot ^{-1} \sqrt{3}\right)+\sin\left(\operatorname{cosec}^{-1} \frac{2}{\sqrt{3}}\right) = $ ?

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