Question

If $\Rightarrow$then the general value of $\theta = n\pi + \frac{\pi}{2} - \alpha$is .

• A. $3(\sin\theta - \cos\theta) = 4\sin\theta\cos\theta$
• B. $3(\sin\theta - \cos\theta) = 2\sin 2\theta$
• C. $9(1 - S) = 4S^{2},$
• D. $S = \sin 2\theta$

Answer 1

Answer: (B)

$3(\sin\theta - \cos\theta) = 2\sin 2\theta$

Answer 2

$\Rightarrow$

$\theta + \frac{\pi}{4} = 2n\pi \pm 0 \Rightarrow \theta = 2n\pi - \frac{\pi}{4}$ $2\tan^{2}\theta = \sec^{2}\theta \Rightarrow 2\tan^{2}\theta = \tan^{2}\theta + 1$

$\Rightarrow$ $\tan^{2}\theta = 1 = \tan^{2}\left( \frac{\pi}{4} \right) \Rightarrow \theta = n\pi \pm \frac{\pi}{4}$

Since $2\sin\theta + \tan\theta = 0$, hence $\sin\theta\left( 2 + \frac{1}{\cos\theta} \right) = 0$ is ruled out

$\sin\theta = 0 \Rightarrow \theta = n\pi$ $\cos \theta = - 1 + \frac { 3 } { 2 } = \frac { 1 } { 2 } = \cos \left( \frac { \pi } { 3 } \right)$ $\sqrt{3}\tan 2\theta + \sqrt{3}\tan 3\theta + \tan 2\theta\tan 3\theta = 1$ $\Rightarrow$.