Q62 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics
If $(\sin \theta-\operatorname{cosec} \theta)^2+(\cos \theta+\sec \theta)^2=5$ and $\theta$ lies in the third quadrant, then $(\sin \theta+\cos \theta)^3=$
- Correct answerA$-2 \sqrt{2}$
- B$2 \sqrt{2}$
- C$4$
- D$-4$
Solution
$\begin{gathered}(\sin \theta-\operatorname{cosec} \theta)^2+(\cos \theta+\sec \theta)^2=5 \\ \sin ^2 \theta+\operatorname{cosec}^2 \theta-2+\cos ^2 \theta+\sec ^2 \theta+2=5 \\ 1+\cot ^2 \theta+1+\tan ^2 \theta=4 \Rightarrow \tan ^2 \theta+\frac{1}{\tan ^2 \theta}=2 \\ \Rightarrow \tan ^2 \theta=1 \Rightarrow \tan \theta=1 \Rightarrow \theta=\pi+\frac{\pi}{4}=\frac{5 \pi}{4} \\ \sin \theta=\sin \left(\frac{5 \pi}{4}\right)=\frac{-1}{\sqrt{2}} ; \cos \theta=\cos \left(\frac{5 \pi}{4}\right)=\frac{-1}{\sqrt{2}} \\ \sin \theta+\cos \theta=-\sqrt{2} \Rightarrow(\sin \theta+\cos \theta)^3=-2 \sqrt{2}\end{gathered}$
Hence, the correct answer is option: 'A'.