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  • Can someone help me with this, - Trigonometry - BITSAT-2

The value of Cot^{-1}\frac{3}{4}+Sin^{-1}\frac{5}{13}     is

  • Option 1)

    Sin^{-1}\, \, \frac{12}{13}

  • Option 2)

    Sin^{-1}\, \, \frac{63}{65}

  • Option 3)

    Sin^{-1}\, \, \frac{5}{12}

  • Option 4)

    Sin^{-1}\, \, \frac{65}{68}

 

Answers (1)

best_answer

As we learnt in

Formulae of Inverse Trigonometric Functions -

\sin ^{-1}x + \sin ^{-1}y= \sin ^{-1}\left [ x\sqrt{1-y^{2}}+ y \sqrt{1-x^{2}} \right ]

\sin ^{-1}x + \sin ^{-1}y=\pi - \sin ^{-1}\left [ x\sqrt{1-y^{2}}+ y \sqrt{1-x^{2}} \right ]

- wherein

x\geqslant 0, y\geqslant 0, x^{2}+ y^{2}\leqslant 1

 

x\geqslant 0, y\geqslant 0, x^{2}+y^{2}> 1

 

 or 

 

Formulae of Inverse Trigonometric Functions -

\cos ^{-1}x + \cos ^{-1}y = \cos ^{-1}\left \left ( xy- \sqrt{1-x^{2}}\sqrt{1-y^{2}} \right )

- wherein

x\geqslant 0, y\geqslant 0

 

 \cot ^{-1}\frac{3}{4}+ \sin -1\frac{s}{13}

\Rightarrow \tan ^{-1}\frac{4}{3}+ \tan -1\frac{5}{12}

\Rightarrow \tan ^{-1} \frac{\frac{4}{3}+\frac{s}{12}}{1-\frac{20}{36}}= \tan -1 \frac{63}{16}

\therefore \sin^{-1}(\frac{63}{65})

 

 


Option 1)

Sin^{-1}\, \, \frac{12}{13}

Incorrect option

Option 2)

Sin^{-1}\, \, \frac{63}{65}

correct option

Option 3)

Sin^{-1}\, \, \frac{5}{12}

Incorrect option

Option 4)

Sin^{-1}\, \, \frac{65}{68}

Incorrect option

Posted by

Plabita

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