If tan beta = frac{n sin alpha cos alpha}{1 - | vedclass

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(A) Given $	an \beta = \frac{n \sin \alpha \cos \alpha}{1 - n \cos^2 \alpha}$.Dividing numerator and denominator by $\cos^2 \alpha$,we get $	an \bet

Vedclass · Accepted Answer

$\frac{-1}{n-1}$

Vedclass · Answer

(A) Given $	an \beta = \frac{n \sin \alpha \cos \alpha}{1 - n \cos^2 \alpha}$.Dividing numerator and denominator by $\cos^2 \alpha$,we get $	an \beta = \frac{n 	an \alpha}{\sec^2 \alpha - n} = \frac{n 	an \alpha}{1 + 	an^2 \alpha - n} = \frac{n 	an \alpha}{(1 - n) + 	an^2 \alpha}$.Now,$	an (\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$.Substituting $	an \beta$:$	an (\alpha + \beta) = \frac{	an \alpha + \frac{n 	an \alpha}{(1 - n) + 	an^2 \alpha}}{1 - 	an \alpha \left( \frac{n 	an \alpha}{(1 - n) + 	an^2 \alpha} ight)}$$= \frac{	an \alpha ((1 - n) + 	an^2 \alpha + n)}{(1 - n) + 	an^2 \alpha - n 	an^2 \alpha} = \frac{	an \alpha (1 + 	an^2 \alpha)}{(1 - n)(1 + 	an^2 \alpha)} = \frac{	an \alpha}{1 - n}$.Therefore,$	an (\alpha + \beta) \cdot \cot \alpha = \left( \frac{	an \alpha}{1 - n} ight) \cdot \frac{1}{	an \alpha} = \frac{1}{1 - n} = \frac{-1}{n - 1}$.

If $\tan \beta = \frac{n \sin \alpha \cos \alpha}{1 - n \cos^2 \alpha}$,then $\tan (\alpha + \beta) \cdot \cot \alpha =$

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