Q62 / 160 · AP EAMCET 2023 (15 May Shift 2) · Mathematics

$\begin{aligned} & \text { If }\left[1-\cos \left(\frac{\pi}{2}+\alpha\right)+\sin \left(\frac{3 \pi}{2}+\alpha\right)\right]^2+ \\ & {\left[1-\sin \left(\frac{3 \pi}{2}-\alpha\right)-\cos \left(\frac{3 \pi}{2}+\alpha\right)\right]^2=a+b \sin ^2\left(\frac{\pi}{4}+\alpha\right)} \end{aligned}$ then $a^2+b^2=$

  • A
    $20$
  • Correct answer
    B
    $52$
  • C
    $40$
  • D
    $32$

Solution

$\begin{aligned} & \text {Since }\left[1-\cos \left(\frac{\pi}{2}+\alpha\right)+\sin \left(\frac{3 \pi}{2}+\alpha\right)\right]^2 \\ & +\left[1-\sin \left(\frac{3 \pi}{2}-\alpha\right)-\cos \left(\frac{3 \pi}{2}+\alpha\right)\right]^2 \\ & =a+b \sin ^2\left(\frac{\pi}{4}+\alpha\right) \\ & \Rightarrow(1+\sin \alpha-\cos \alpha)^2+(1+\cos \alpha-\sin \alpha)^2 \\ & =a+b \sin ^2\left(\frac{\pi}{4}+\alpha\right) \\ & \Rightarrow 1+\sin ^2 \alpha+\cos ^2 \alpha+2 \sin \alpha-2 \sin \alpha \cos \alpha-2 \cos \alpha \\ & +1+\cos ^2 \alpha+\sin ^2 \alpha+2 \cos \alpha-2 \sin \alpha \cos \alpha-2 \sin \alpha \\ & =a+b \sin ^2\left(\frac{\pi}{4}+\alpha\right) \\ & \Rightarrow 4-4 \sin \alpha \cos \alpha=a+b\left(\sin \left(\frac{\pi}{4}\right) \cos \alpha+\cos \frac{\pi}{4} \sin \alpha\right)^2 \\ & \Rightarrow 4-4 \sin \alpha \cos \alpha=a+\frac{b}{2}(\cos \alpha+\sin \alpha)^2 \\ & \Rightarrow 4-4 \sin \alpha \cos \alpha=a+\frac{b}{2}(1+2 \sin \alpha \cos \alpha) \\ & \Rightarrow 4-4 \sin \alpha \cos \alpha=a+\frac{b}{2}+b \sin \alpha \cos \alpha \\ & \Rightarrow b=-4, a+\frac{b}{2}=4 \Rightarrow a=4+2=6 \\ & \end{aligned}$ Now, $a^2+b^2=16+36=52$

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

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