Question

If $\tan \alpha =\sqrt{a}$, where $a$ is a rational number which is not a perfect square, then which of the following is a rational number ?

• A. $\sin 2\alpha$
• B. $\tan 2\alpha$
• C. $\cos 2\alpha$
• D. $\text{cosec}2\alpha$

Answer 1

Answer: (C)

$\cos 2\alpha$

Let $\tan \alpha =\sqrt{a}$.......(1).

i) $\sin 2\alpha =\frac{2\tan \alpha }{1+{\tan }^2\alpha }=\frac{2\sqrt{a}}{1+a}$ [Using (1)],

which is not a rational number due to the presence of the square root, as $a$ is not perfect square.

ii) $\tan 2\alpha =\frac{2\tan \alpha }{1-{\tan }^2\alpha }=\frac{2\sqrt{a}}{1-a}$ [Using (1)],

which is again not a rational number due to the previous reason.

iii) $\cos 2\alpha =\frac{1-{\tan }^2\alpha }{1+{\tan }^2\alpha }=\frac{1-a}{1+a}$ [Using (1)],

no square root is present, so it is going to be the rational one.

iv) $\text{cosec}2\alpha =\frac{1-a}{2\sqrt{a}}$ [Using (i)],

which is not a rational number.

So, option (C) is correct.