frac{sin 81^o + cos 81^o}{sin 81^o - cos 81^o | vedclass

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Question

(D) Given expression $E = \frac{\sin 81^o + \cos 81^o}{\sin 81^o - \cos 81^o}$.Divide numerator and denominator by $\cos 81^o$:$E = \frac{	an 81^o +

Vedclass · Accepted Answer

$	an 54^o$

Vedclass · Answer

(D) Given expression $E = \frac{\sin 81^o + \cos 81^o}{\sin 81^o - \cos 81^o}$.Divide numerator and denominator by $\cos 81^o$:$E = \frac{	an 81^o + 1}{	an 81^o - 1}$.Multiply numerator and denominator by $-1$:$E = \frac{1 + 	an 81^o}{1 - 	an 81^o}$.Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,where $A = 45^o$ and $B = 81^o$:$E = 	an(45^o + 81^o) = 	an 126^o$.Alternatively,note that $\sin 81^o = \cos 9^o$ and $\cos 81^o = \sin 9^o$:$E = \frac{\cos 9^o + \sin 9^o}{\cos 9^o - \sin 9^o} = \frac{1 + 	an 9^o}{1 - 	an 9^o} = 	an(45^o + 9^o) = 	an 54^o$.

$\frac{\sin 81^o + \cos 81^o}{\sin 81^o - \cos 81^o}$ is equal to

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