If sinθ + cosθ = p and secθ + cosecθ = q, then prove that q (p2

Name: If sinθ + cosθ = p and secθ + cosecθ = q, then prove that q (p2 – 1) = 2p.
Uploaded: Mon, 2021-02-01 16:43
Description: If sinθ + cosθ = p and secθ + cosecθ = q, then prove that q (p2 – 1) = 2p.

If sinθ + cosθ = p and secθ + cosecθ = q, then prove that q (p² – 1) = 2p.

Answers (1)

Solution. Given :-sinθ + cosθ = p
and secθ + cosecθ = q
To prove :-q (p² – 1) = 2p
Taking left hand side
q.(p²– 1) =
Put value of q and p we get
$\left ( \sec \theta +\cos ec\theta \right )\left [ \left ( \sin \theta +\cos \theta \right )^{2}-1 \right ]$
$\left ( \because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab\right )$
= $\left ( \frac{1}{\cos \theta +\frac{1}{\sin \theta }} \right )\left [ \left ( \sin^{2} \theta+\cos^{2} \theta+2 \sin \theta \cos \theta \right ) -1\right ]$
$\left ( \because \sec \theta = \frac{1}{\cos \theta } ,\cos ec\theta =\frac{1}{\sin \theta } \right )$
$= \frac{1}{\cos \theta }\left [ \sin ^{2}\theta +\cos ^{2}\theta +2\sin \theta \cos \theta -1 \right ]$
$+ \frac{1}{\sin \theta }\left [ \sin ^{2}\theta +\cos ^{2}\theta +2\sin \theta \cdot \cos \theta -1 \right ]$
$= \frac{\sin ^{2}}{\cos \theta }+\cos \theta+2\sin \theta -\frac{1}{\cos \theta}+\sin \theta+\frac{\cos^{2} \theta}{\sin \theta}+2\cos \theta-\frac{1}{\sin \theta}$
$= 3\cos \theta +3\sin \theta -\frac{1}{\cos \theta}+\frac{\left ( 1+\cos ^{2}\theta \right )}{\cos \theta }+\frac{\left ( 1+\sin ^{2}\theta \right )}{\sin \theta }-\frac{1}{\sin \theta }$
$\left ( \because \sin ^{2}\theta = 1-\cos ^{2}\theta \right )$
$\left ( \because \cos ^{2}\theta = 1-\sin ^{2}\theta \right )$
$= 3\cos \theta +3\sin \theta+\frac{1}{\cos \theta}\times \left ( -1+1-\cos^{2} \theta \right )+\frac{1}{\sin \theta }\times \left ( 1-\sin^{2} \theta-1 \right )$
$\left ( \because \sin^{2} \theta+\cos^{2} \theta= 1 \right )$
$= 3\cos \theta -\cos \theta +3\sin \theta -\sin \theta$
$= 2\cos \theta +2\sin \theta$
$= 2\left ( \cos \theta +\sin \theta \right )$
2p (R.H.S)
Hence proved.