જો sin alpha = frac{2pq}{p^2 + q^2} હોય,તો se | vedclass

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(A) આપેલ છે કે $\sin \alpha = \frac{2pq}{p^2 + q^2}$.આપણે જાણીએ છીએ કે $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{4p^2q^2}{(p^2 + q^2)^2} = \frac{

Vedclass · Accepted Answer

$\frac{p-q}{p+q}$

Vedclass · Answer

(A) આપેલ છે કે $\sin \alpha = \frac{2pq}{p^2 + q^2}$.આપણે જાણીએ છીએ કે $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{4p^2q^2}{(p^2 + q^2)^2} = \frac{(p^2 + q^2)^2 - 4p^2q^2}{(p^2 + q^2)^2} = \frac{(p^2 - q^2)^2}{(p^2 + q^2)^2}$.તેથી,$\cos \alpha = \frac{p^2 - q^2}{p^2 + q^2}$.હવે,$\sec \alpha = \frac{1}{\cos \alpha} = \frac{p^2 + q^2}{p^2 - q^2}$ અને $	an \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{2pq}{p^2 - q^2}$.હવે,$\sec \alpha - 	an \alpha = \frac{p^2 + q^2}{p^2 - q^2} - \frac{2pq}{p^2 - q^2} = \frac{p^2 + q^2 - 2pq}{p^2 - q^2} = \frac{(p-q)^2}{(p-q)(p+q)} = \frac{p-q}{p+q}$.

જો $\sin \alpha = \frac{2pq}{p^2 + q^2}$ હોય,તો $\sec \alpha - \tan \alpha = $

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