Question

if $\sin \theta +\cos \theta =p$ and $\sec \theta +cosec\theta =q$
Then, $q(p^2-1)=$

• A. 2
• B. 2p
• C. $\frac{2}{P^2}$
• D. None

Answer 1

The correct option is B 2p

Given: $\sin \theta +\cos \theta =p$ and $\sec \theta +\cos \sec \theta =q$
Now, $q(p^2-1)=(sec\theta +cosec\theta )\left(\right(sin\theta +cos\theta {)}^2-1)$
= $(sec\theta +cosec\theta )(si{n}^2\theta +co{s}^2\theta +2sin\theta \cos \theta -1)$
= $(\frac{1}{cos\theta }+\frac{1}{\sin \theta })(1+2sin\theta \cos \theta -1)$
= $\left(\frac{cos\theta +\sin \theta }{cos\theta \sin \theta }\right)\left(2sin\theta \cos \theta \right)$
= $2(cos\theta +\sin \theta )$
= $2p$