Q187 / 270 · NDA 2018 (Phase 2) · Mathematics

If $\sec (\theta-\alpha), \sec \theta$ and $\sec (\theta+\alpha)$ are in AP, where $\cos$ $\alpha \neq 1$, then what is the value of $\sin ^2 \theta+\cos \alpha$ ?

  • Correct answer
    A
    0
  • B
    1
  • C
    -1
  • D
    1/2

Solution

$\begin{aligned} & \text { } \frac{2}{\cos \theta}=\frac{\cos (\theta+\alpha)+\cos (\theta-\alpha)}{\cos (\theta+\alpha) \cos (\theta-\alpha)}=\frac{2 \cos \theta \cdot \cos \alpha}{\cos ^2 \theta-\sin ^2 \alpha} \\ & \Rightarrow \cos ^2 \theta \cos \alpha=\cos ^2 \theta-\sin ^2 \alpha \\ & \Rightarrow \sin ^2 \alpha=\cos ^2 \theta(1-\cos \alpha) \\ & \Rightarrow \cos ^2 \theta=\frac{\sin ^2 \alpha}{1-\cos \alpha}=1+\cos \alpha \\ & \Rightarrow 1-\sin ^2 \theta=1+\cos \alpha \\ & \Rightarrow \sin ^2 \theta+\cos \alpha=0\end{aligned}$

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

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