Question

If $\sin \theta+\cos \theta=p$ and $\sin ^{3} \theta+\cos ^{3} \theta=q$ then $p\left(p^{2}-3\right)$ is equal to

• A. q
• B. 2 q
• C. -q
• D. -2 q

Answer 1

Answer: (D)

Given, $\sin \theta+\cos \theta=p$ ...(i)
and $\sin ^{3} \theta+\cos ^{3} \theta=q$ ...(ii)
$\Rightarrow (\sin \theta+\cos \theta)$
$\left(\sin ^{2} \theta-\sin \theta \cdot \cos \theta+\cos ^{2} \theta\right)=q$
$\Rightarrow p(1-\sin \theta \cdot \cos \theta)=q$
[From Eq. (i) and $\left.\sin ^{2} \theta+\cos ^{2} \theta=1\right]$
$\Rightarrow 1-\sin \theta \cdot \cos \theta=\frac{q}{p}$
$\Rightarrow \sin \theta \cdot \cos \theta=1-\frac{q}{p}$ ...(iii)
On squaring both sides of Eq. (i), we get
$\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cdot \cos \theta=p^{2}$
$\Rightarrow 1+2\left(1-\frac{q}{p}\right)=p^{2}$ [from Eq. (iii)]
$\Rightarrow p+2(p-q)=p^{3}$
$\Rightarrow 3 p-2 q=p^{3}$
$\Rightarrow p^{3}-3 p=-2 q$
$\Rightarrow p\left(p^{2}-3\right)=-2 q$