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AP EAMCET Engineering 2018 Apr 22 Shift 2 Paper

Section: Mathematics

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Question : 20 of 160

Marks: +1, -0

If

\tan\left(\frac{\pi}{4}+\frac{y}{2}\right)=\tan^3\left(\frac{\pi}{4}+\frac{x}{2}\right),

\frac{3\sin x+\sin^3 x}{1+3\sin^2 x}=

0
1
sin 2y
sin y

Solution:

Given:

\tan\left(\frac{\pi}{4}+\frac{y}{2}\right)=\tan^3\left(\frac{\pi}{4}+\frac{x}{2}\right) ,

\frac{1+\tan\frac{y}{2}}{1-\tan\frac{y}{2}}=\left(\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}\right)^3

\frac{\cos\frac{y}{2}+\sin\frac{y}{2}}{\cos\frac{y}{2}-\sin\frac{y}{2}}=\left(\frac{\cos\frac{x}{2}+\sin\frac{x}{2}}{\cos\frac{x}{2}-\sin\frac{x}{2}}\right)^3

Take square both sides,

\frac{1+\sin y}{1-\sin y}=\left(\frac{1+\sin x}{1-\sin x}\right)^3

Apply component and dividend rule,

\frac{2}{2\sin y}=\frac{(1+\sin x)^3+(1-\sin x)^3}{(1+\sin x)^3-(1-\sin x)^3}

\frac{3\sin x+\sin^3 x}{1+3\sin^2 x}=\sin y

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