Question

Question 10
If $sin\theta +cos\theta =p$ and $sec\theta +cosec\theta =q$, then prove that $q(p^2-1)=2p$.

Answer 1

Given that, $sin\theta +cos\theta =p\dots \dots \left(i\right)$

And, $sec\theta +cosec\theta =q$

$\Rightarrow \frac{1}{cos\theta }+\frac{1}{sin\theta }=q[\because sec\theta =\frac{1}{cos\theta }andcosec\theta =\frac{1}{sin\theta }]$

$\Rightarrow \frac{sin\theta +cos\theta }{sin\theta .cos\theta }=q$

$\Rightarrow \frac{p}{sin\theta .cos\theta }=q$ [from Eq. (i)]

$\Rightarrow sin\theta .cos\theta =\frac{p}{q}$ [from Eq. (i)...(ii)]

$sin\theta +cos\theta =p$

On squaring both sides, we get;

$(sin\theta +cos\theta {)}^2=p^2$

$\Rightarrow (si{n}^2\theta +co{s}^2\theta )+2sin\theta .cos\theta =p^2[\because (a+b{)}^2=a^2+2ab+b^2]$]

$\Rightarrow 1+2sin\theta .cos\theta =p^2[\because si{n}^2\theta +co{s}^2\theta =1]$

$\Rightarrow 1+2.\frac{p}{q}=p^2$ [from Eq. (ii)]

$\Rightarrow q+2p=p^{2}q\Rightarrow 2p=p^{2}q-q$

$\Rightarrow q(p^2-1)=2p$