If sin x + sin 2 x = 1 , x ( 0 , 2 ) , then ( cos 12 x + tan 12 x ) + 3 ( cos 10 x + tan 10 x + cos 8 x + tan 8 x ) + ( cos 6 x + tan 6 x ) is equal to : [2025]

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Solution

Q.
If $\sin x + \sin^{2} x = 1, x \in (0, \frac{π}{2})$ , then $(\cos^{12} x + \tan^{12} x) + 3 (\cos^{10} x + \tan^{10} x + \cos^{8} x + \tan^{8} x) + (\cos^{6} x + \tan^{6} x)$ is equal to : [2025]

1 2

2 4

3 3

4 1

Ans.
(1)

We have, $\sin x + \sin^{2} x = 1$

$\Rightarrow \sin x = 1 - \sin^{2} x \Rightarrow \sin x = \cos^{2} x \Rightarrow \tan x = \cos x$

Now, $(\cos^{12} x + \tan^{12} x) + 3 (\cos^{10} x + \tan^{10} x + \cos^{8} x + \tan^{8} x) + (\cos^{6} x + \tan^{6} x)$

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

$= 2 \cos^{12} x + 6 \cos^{10} x + 6 \cos^{8} x + 2 \cos^{6} x$

$= 2 \sin^{6} x + 6 \sin^{5} x + 6 \sin^{4} x + 2 \sin^{3} x$ [ $∵ \sin x = \cos^{2} x$ ]

$= 2 \sin^{3} x [\sin^{3} x + 3 \sin^{2} x + 3 \sin x + 1]$

$= 2 \sin^{3} x [{(1 + \sin x)}^{3}]$

$= 2 [{(\sin x + \sin^{2} x)}^{3}] = 2$ { $∵ \sin x + \sin^{2} x = 1$ }

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