The correct option is C 176 We have #160; cos^ 145+tan^ 123 =tan^ 134+tan^ 123 =tan^ 1(34+231 34.23) =tan^ 1(171212) =tan^ 1(176) ...

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Sin(A+B)Sin(A-B)

tan cos-145+t...

Question

$tan ({cos}^{- 1} \frac{4}{5} + {tan}^{- 1} \frac{2}{3}) =$

\frac{6}{17}

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\frac{- 17}{6}

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\frac{17}{6}

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\frac{16}{7}

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Solution

The correct option is C $\frac{17}{6}$
We have
${cos}^{- 1} \frac{4}{5} + {tan}^{- 1} \frac{2}{3}$

$= {tan}^{- 1} \frac{3}{4} + {tan}^{- 1} \frac{2}{3}$

$= {tan}^{- 1} ⎛ ⎜ ⎜ ⎜ ⎝ \frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{3}{4} . \frac{2}{3}} ⎞ ⎟ ⎟ ⎟ ⎠$

$= {tan}^{- 1} ⎛ ⎜ ⎜ ⎜ ⎝ \frac{\frac{17}{12}}{\frac{1}{2}} ⎞ ⎟ ⎟ ⎟ ⎠$

$= {tan}^{- 1} (\frac{17}{6})$
Now,
$tan ⌊ {cos}^{- 1} \frac{4}{5} + {tan}^{- 1} \frac{2}{3} ⌋$
$= tan ⌊ {tan}^{- 1} (\frac{17}{6}) ⌋$
$= \frac{17}{6}$

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tan(cos−145+tan−123)=

The correct option is C 176We have cos−145+tan−123=tan−134+tan−123=tan−1⎛⎜ ⎜ ⎜⎝34+231−34.23⎞⎟ ⎟ ⎟⎠=tan−1⎛⎜ ⎜ ⎜⎝171212⎞⎟ ⎟ ⎟⎠=tan−1(176)Now,tan⌊cos−145+tan−123⌋=tan⌊tan−1(176)⌋=176

$tan ({cos}^{- 1} \frac{4}{5} + {tan}^{- 1} \frac{2}{3}) =$