If sin alpha = frac{2pq}{p^2 + q^2},then sec | vedclass

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Question

(A) Given $\sin \alpha = \frac{2pq}{p^2 + q^2}$.We know that $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{4p^2q^2}{(p^2 + q^2)^2} = \frac{(p^2 + q^2

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

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(A) Given $\sin \alpha = \frac{2pq}{p^2 + q^2}$.We know that $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{4p^2q^2}{(p^2 + q^2)^2} = \frac{(p^2 + q^2)^2 - 4p^2q^2}{(p^2 + q^2)^2} = \frac{(p^2 - q^2)^2}{(p^2 + q^2)^2}$.Thus,$\cos \alpha = \frac{p^2 - q^2}{p^2 + q^2}$.Then,$\sec \alpha = \frac{1}{\cos \alpha} = \frac{p^2 + q^2}{p^2 - q^2}$ and $	an \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{2pq}{p^2 - q^2}$.Now,$\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}$.

If $\sin \alpha = \frac{2pq}{p^2 + q^2}$,then $\sec \alpha - \tan \alpha = $

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