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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 = $

Similar Questions

$\sin 4\theta$ can be written as

$\tan 7 \frac{1}{2}^{\circ}$ is equal to

Which one of the following is true?

If $\sec \theta + \tan \theta = m$ $(m > 1)$,then the value of $\sin \theta$ is $(0^{\circ} < \theta < 90^{\circ})$.

If $\tan A = 2\tan B + \cot B,$ then $2\tan (A - B) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module