If $\tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}, 0 \leq \alpha \leq \frac{\pi}{2}$,…

If $\tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}, 0 \leq \alpha \leq \frac{\pi}{2}$, then the value of $\cos 2 \theta$ is
  1. $\cos 2 \alpha$
  2. $\sin \alpha$
  3. $\cos \alpha$
  4. $\sin 2 \alpha$

Solution

$\begin{array}{ll} & \tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha} \\ & \frac{\sin \theta}{\cos \theta}=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha} \\ \therefore \quad & \sin \alpha \sin \theta+\cos \alpha \sin \theta=\sin \alpha \cos \theta-\cos \alpha \cos \theta \\ \therefore \quad & \cos \alpha \cos \theta+\sin \alpha \sin \theta=\sin \alpha \cos \theta-\cos \alpha \sin \theta \\ \therefore \quad & \cos (\alpha-\theta)=\sin (\alpha-\theta) \\ \therefore \quad & \alpha-\theta=\frac{\pi}{4} \\ \therefore \quad & \theta=\alpha-\frac{\pi}{4} \\ \therefore \quad & 2 \theta=2 \alpha-\frac{\pi}{2} \\ \therefore \quad & \cos 2 \theta=\cos \left(2 \alpha-\frac{\pi}{2}\right) \\ & =\cos \left[\because 0 \leq \alpha \leq \frac{\pi}{2}\right] \\ & \left.=\cos \left(\frac{\pi}{2}-2 \alpha\right)\right] \quad \ldots[\because \cos (-\theta)=\cos \theta] \\ \therefore \quad & \cos 2 \theta=\sin 2 \alpha\end{array}$

Asked in: MHT CET 2023 (12 May Shift 1)

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