Question

If $\theta = n\pi + ( - 1)^{n}\frac{\pi}{3}$, then the

general value of $\cos^{2}(A + B) + \sin^{2}(A + B) + \cos 2B = 0$is.

• A. $1 + \cos 2B = 0$
• B. $\cos 2B = \cos\pi$
• C. $2B = 2n\pi + \pi$
• D. $B = (2n + 1)\frac{\pi}{2},n \in Z.$

Answer 1

Answer: (B)

$\cos 2B = \cos\pi$

Answer 2

$\tan^{2}\theta - 2\tan\theta + 1 = 0$

$\Rightarrow$

= 0, (from the given condition)

$\tan\theta = 1 = \tan\frac{\pi}{4} \Rightarrow$.

Trick : In such type of problems, the general value of $\theta = n\pi + \frac{\pi}{4}$ is given by $2\cos^{2}\theta - (\sqrt{2} + 1)\cos\theta - 1 + \frac{(\sqrt{2} + 1)}{\sqrt{2}} = 0$. So the general value of $\Rightarrow$ is

$\Rightarrow$.