AP EAMCET Engineering 23 Apr 2019 Shift 1 Paper

Section: Mathematics

Question : 25 of 160

Marks: +1, -0

If -

\frac{\pi}{2}

\frac{\pi}{2}

\left(\tan\left(\frac{\pi}{4}+\frac{\theta}{2}\right)\right)

Solution:

Consider the expression,

x=\log \tan\left(\frac{\pi}{4}+\frac{\theta}{2}\right)

Then,

\tan\left(\frac{\pi}{4}+\frac{\theta}{2}\right)=e^x

\frac{1+\tan\frac{\theta}{2}}{1-\tan\frac{\theta}{2}}=e^{x}

This implies,

\frac{e^{y}-1}{e^{y}+1}=\tan\frac{x}{2}

\tanh\left(\frac{y}{2}\right)=\tan\left(\frac{x}{2}\right)

By the componendo and dividendo rule,

\frac{2\tan\frac{\theta}{2}}{2}=\frac{e^{x}-1}{e^{x}+1}

\tan\frac{\theta}{2}=\frac{e^{\frac{x}{2}}-e^{\frac{x}{2}}}{e^{\frac{x}{2}}+e^{\frac{x}{2}}}

=\tanh\left(\frac{x}{2}\right)

This implies,

\frac{x}{2}=\tanh^{-1}\left(\tan\frac{\theta}{2}\right)

x=2\tanh^{-1}\left(\tan\frac{\theta}{2}\right)

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