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TS EAMCET 5 May 2018 Shift 1 Solved Paper

Section: Mathematics

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Question : 64 of 160

Marks: +1, -0

\sin^{-1}\left(\frac{12}{13}\right)

\cos^{-1}\left(\frac{4}{5}\right)

\tan^{-1}\left(\frac{63}{16}\right)

=

2π
π
0
- π

Solution:

The given expression is,

\sin^{-1}\left(\frac{12}{13}\right)

+

\cos^{-1}\left(\frac{4}{5}\right)

+

\tan^{-1}\left(\frac{63}{16}\right)

…… (1)

Assume,

\theta=\sin^{-1}\left(\frac{12}{13}\right)

\sin\theta=\frac{12}{13}

Consider the figure,

\tan\theta=\frac{12}{5}

\theta=\tan^{-1}\left(\frac{12}{5}\right)

Assume,

\phi=\cos^{-1}\left(\frac{4}{5}\right)

\cos\phi=\frac{4}{5}

Consider the figure,

\tan\phi=\frac{3}{4}

\phi=\tan^{-1}\left(\frac{3}{4}\right)

Substitute the value in equation (1),

\tan^{-1}\left(\frac{12}{5}\right)+\tan^{-1}\left(\frac{3}{4}\right)+\tan^{-1}\left(\frac{63}{16}\right)

From,

\tan^{-1}A+\tan^{-1}B=\pi+\tan^{-1}\left(\frac{A+B}{1-AB}\right)

the above equation is simplified as,

\tan^{-1}\left(\frac{12}{5}\right)+\tan^{-1}\left(\frac{3}{4}\right)

+\tan^{-1}\left(\frac{63}{16}\right)=\pi

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