Trigonometry – Trigonometric Identities – JEE Main 29 Jan 2025 Shift 2

Solution:

Given $\sin x + \sin^{2}x = 1 \Rightarrow \sin x = 1 – \sin^{2}x = \cos^{2}x$.

Now, $\tan^{2}x = \frac{\sin^{2}x}{\cos^{2}x} = \frac{\sin^{2}x}{\sin x} = \sin x$.

Since $x \in (0, \frac{\pi}{2})$, $\tan x = \sqrt{\sin x} = \cos x$.

Substitute $\tan x = \cos x$ into the given expression $E$:

$E = (\cos^{12}x + \cos^{12}x) + 3(\cos^{10}x + \cos^{10}x + \cos^{8}x + \cos^{8}x) + (\cos^{6}x + \cos^{6}x)$

$E = 2\cos^{12}x + 3(2\cos^{10}x + 2\cos^{8}x) + 2\cos^{6}x$

$E = 2[\cos^{12}x + 3\cos^{10}x + 3\cos^{8}x + \cos^{6}x]$

The term inside the bracket is the expansion of $(\cos^{4}x + \cos^{2}x)^{3}$:

$E = 2[(\cos^{4}x + \cos^{2}x)^{3}]$

Substitute $\cos^{2}x = \sin x$ (so $\cos^{4}x = \sin^{2}x$):

$E = 2[(\sin^{2}x + \sin x)^{3}]$

Since $\sin^{2}x + \sin x = 1$:

$E = 2(1)^{3} = 2$

Ans. (3)