The value of tan left(frac{pi}{4}+frac{1}{2} | vedclass

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(B) Let $	heta = \frac{1}{2} \cos ^{-1}\left(\frac{a}{b}ight)$. Then $2	heta = \cos ^{-1}\left(\frac{a}{b}ight)$,which implies $\cos 2	heta = \

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$\frac{2 b}{a}$

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(B) Let $	heta = \frac{1}{2} \cos ^{-1}\left(\frac{a}{b}ight)$. Then $2	heta = \cos ^{-1}\left(\frac{a}{b}ight)$,which implies $\cos 2	heta = \frac{a}{b}$. The given expression is $	an \left(\frac{\pi}{4}+	hetaight)+	an \left(\frac{\pi}{4}-	hetaight)$. Using the formula $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$ and $	an(A-B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$,where $	an(\frac{\pi}{4}) = 1$: $= \frac{1+	an 	heta}{1-	an 	heta} + \frac{1-	an 	heta}{1+	an 	heta}$ $= \frac{(1+	an 	heta)^2 + (1-	an 	heta)^2}{(1-	an 	heta)(1+	an 	heta)}$ $= \frac{1 + 2	an 	heta + 	an^2 	heta + 1 - 2	an 	heta + 	an^2 	heta}{1 - 	an^2 	heta}$ $= \frac{2(1+	an^2 	heta)}{1-	an^2 	heta}$ $= \frac{2}{\frac{1-	an^2 	heta}{1+	an^2 	heta}}$ $= \frac{2}{\cos 2	heta}$ Substituting $\cos 2	heta = \frac{a}{b}$,we get: $= \frac{2}{a/b} = \frac{2b}{a}$.

The value of $\tan \left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right)+\tan \left(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1}\left(\frac{a}{b}\right)\right)$ is

Similar Questions

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$I$. $\sin ^{-1}(y^2-4y+6)+\cos ^{-1}(y^2-4y+6) = \frac{\pi}{2}, \forall y \in R$
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