Question

If $\sqrt{x} + \frac{1}{\sqrt{x}} = 2\cos\theta,$ then $\tan(A + 2B)$

• A. $2\cos 6\theta$
• B. $2\cos 12\theta$
• C. $2\cos 3\theta$
• D. $2\sin 3\theta$

Answer 1

Answer: (B)

$2\cos 12\theta$

Answer 2

Given, $\sqrt{x} + \frac{1}{\sqrt{x}} = 2\cos\theta$ ........(i)

On squaring both sides we get, $x + \frac{1}{x} + 2 = 4\cos^{2}\theta$

⇒ $x + \frac{1}{x} = 4\cos^{2}\theta - 2$

$\Rightarrow x + \frac{1}{x} = 2(2\cos^{2}\theta - 1) = 2\cos 2\theta$ ........(ii)

Again squaring both sides,

$x^{2} + \frac{1}{x^{2}} + 2 = 4\cos^{2}2\theta$

⇒$2\tan A = 3\tan B,$

⇒ $x^{2} + \frac{1}{x^{2}} = 2\cos 4\theta$ ......(iii)

Now taking cube of both sides; $\left( x^{2} + \frac{1}{x^{2}} \right)^{3} = (2\cos 4\theta)^{3}$

⇒$x^{6} + \frac{1}{x^{6}} + 3x^{2}.\frac{1}{x^{2}}\left( x^{2} + \frac{1}{x^{2}} \right) = 8\cos^{3}4\theta$

⇒ $x^{6} + \frac{1}{x^{6}} + 3(2\cos 4\theta)$ = $8\cos^{3}4\theta$

⇒ $x^{6} + \frac{1}{x^{6}} = 8\cos^{3}4\theta - 6\cos 4\theta$

⇒ $x^{6} + \frac{1}{x^{6}} = 2(4\cos^{3}4\theta - 3\cos 4\theta) = 2\cos 3(4\theta) = 2\cos 12\theta$.