Q106 / 160 · TS EAMCET 2023 (12 May Shift 1) · Mathematics

$\sin \alpha+\cos \alpha=m \Rightarrow \sin ^6 \alpha+\cos ^6 \alpha=$

  • A
    $\frac{4+3\left(m^2-1\right)^2}{4}$
  • Correct answer
    B
    $\frac{4-3\left(m^2-1\right)^2}{4}$
  • C
    $\frac{3+4\left(m^2-1\right)^2}{4}$
  • D
    $\frac{4-3\left(m^2+1\right)^2}{4}$

Solution

$\begin{aligned} & \sin \alpha+\cos \alpha=m \\ & \sin ^6 \alpha+\cos ^6 \alpha=\left(\sin ^2 \alpha\right)^3+\left(\cos ^2 \alpha\right)^3 \\ & =\left(\sin ^2 \alpha+\cos ^2 \alpha\right)\left(\sin ^4 \alpha-\sin ^2 \alpha \cos ^2 \alpha+\cos ^4 \alpha\right) \\ & =\left(\left(\sin ^2 \alpha\right)^2+\left(\cos ^2 \alpha\right)^2-\sin ^2 \alpha \cos ^2 \alpha\right) \\ & =\left(\sin ^2 \alpha+\cos ^2 \alpha\right)^2-3 \sin ^2 \alpha \cos ^2 \alpha \\ & =1-\frac{3}{4}(2 \sin \alpha \cos \alpha)^2... (i)) \\ & \because \quad \sin \alpha+\cos \alpha=m \\ & \Rightarrow \quad(\sin \alpha+\cos \alpha)^2=m^2 \\ & \Rightarrow \quad 1+2 \sin \alpha \cos \alpha=m^2 \\ & \Rightarrow \quad 2 \sin \alpha \cos \alpha=m^2-1 \\ & \therefore \quad \text { From (i) } \\ & \sin ^6 \alpha+\cos ^6 \alpha=1-\frac{3}{4}\left(m^2-1\right)^2=\frac{4-3\left(m^2-1\right)^2}{4} .\end{aligned}$

Hence, the correct answer is option: 'B'.

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