If frac{1}{2}left(tan left(frac{pi}{24}right) | vedclass

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(B) We know that $	an 	heta + \cot 	heta = \frac{\sin 	heta}{\cos 	heta} + \frac{\cos 	heta}{\sin 	heta} = \frac{\sin^2 	heta + \cos^2 	heta}

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) We know that $	an 	heta + \cot 	heta = \frac{\sin 	heta}{\cos 	heta} + \frac{\cos 	heta}{\sin 	heta} = \frac{\sin^2 	heta + \cos^2 	heta}{\sin 	heta \cos 	heta} = \frac{1}{\sin 	heta \cos 	heta} = \frac{2}{\sin(2	heta)}$.Substituting $	heta = \frac{\pi}{24}$,we get $	an \frac{\pi}{24} + \cot \frac{\pi}{24} = \frac{2}{\sin(\frac{\pi}{12})}$.Since $\frac{\pi}{12} = 15^\circ$,$\sin(15^\circ) = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3}-1}{2\sqrt{2}}$.Thus,$\frac{1}{2} \left(	an \frac{\pi}{24} + \cot \frac{\pi}{24}ight) = \frac{1}{2} \cdot \frac{2}{\sin(15^\circ)} = \frac{1}{\sin(15^\circ)} = \frac{2\sqrt{2}}{\sqrt{3}-1}$.Rationalizing the denominator: $\frac{2\sqrt{2}(\sqrt{3}+1)}{3-1} = \sqrt{6} + \sqrt{2}$.We are given $\sqrt{a^2+a} + \sqrt{a} = \sqrt{6} + \sqrt{2}$.Comparing the terms,$\sqrt{a} = \sqrt{2} \implies a = 2$.Checking the first term: $\sqrt{2^2+2} = \sqrt{6}$,which matches. Therefore,$a = 2$.

If $\frac{1}{2}\left(\tan \left(\frac{\pi}{24}\right)+\cot \left(\frac{\pi}{24}\right)\right)=\sqrt{a^2+a}+\sqrt{a}$,then $a=$

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