Q90 / 170 · Manipal (MET) 2015 · Mathematics
If $\sin A+\cos A=m$ and $\sin ^3 A+\cos ^3 A=n$ then
- A$m^3-3 m+n=0$
- B$n^3-3 n+2 m=0$
- Correct answerC$m^3-3 m+2 n=0$
- D$m^3+3 m+2 n=0$
Solution
We have,
$\sin A+\cos A=m$ and $\sin ^3 A+\cos ^3 A=n$
Now',
$\sin A+\cos A=m$
$\begin{array}{ll}\Rightarrow \quad & (\sin A+\cos A)^3=m^3 \\ \Rightarrow \quad & \sin ^3 A+\cos ^3 A+3 \sin A \cos A \\ & \\ & (\sin A+\cos A)=m^3\end{array}$
$\Rightarrow \quad n+3 \sin A \cos A m=m^3$ ...(i)
Again,
$\sin A+\cos A=m$
$\Rightarrow \sin ^2 A+\cos ^2 A+2 \sin A \cos A=m^2$
$\Rightarrow \quad \sin A \cos A=\frac{m^2-1}{2}$ ...(ii)
From Eqs. (i) and (ii), we get
$n+3 m \frac{\left(m^2-1\right)}{2}=m^3$
$\begin{array}{ll}\Rightarrow & 2 n+3 m^3-3 m=2 m^3 \\ \Rightarrow & m^3-3 m+2 n=0\end{array}$
Hence, the correct answer is option: 'C'.