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# Engineering

If \alpha = \cos ^{-1}\left ( \frac{3}{5} \right ),\beta = \tan ^{-1}\left ( \frac{1}{3} \right ), where 0<\alpha ,\beta <\frac{\pi}{2}, then \alpha -\beta is equal to :
 

  • Option 1)

    \sin ^{-1}\left ( \frac{9}{5\sqrt{10}} \right )
     

  • Option 2)

    \tan ^{-1}\left ( \frac{9}{5\sqrt{10}} \right )

  • Option 3)

    \cos ^{-1}\left ( \frac{9}{5\sqrt{10}} \right )

     

  • Option 4)

    \tan ^{-1}\left ( \frac{9}{14} \right )

 

Answers (1)
S solutionqc

\alpha=\cos^{-1}\frac{3}{5}

\beta =\tan^{-1}\left ( \frac{1}{3} \right )\; \; \; 0<\alpha ,\beta <\frac{\pi}{2}

Then

\cos\: \alpha=\frac{3}{5}, \tan \alpha =\frac{4}{3},\alpha =\tan ^{-1}\left ( \frac{4}{3} \right ),

\tan \beta =\frac{1}{3}                        

\alpha -\beta =\cos^{-1}\left ( \frac{3}{5} \right )-\tan^{-1}\left ( \frac{1}{3} \right )

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

               

               =\tan^{-1}\left ( \frac{1}{1+\frac{4}{9}}\right )=\tan^{-1}\left ( \frac{9}{13} \right )

               \tan^{-1}\left ( \frac{9}{13} \right )

No option matched.


Option 1)

\sin ^{-1}\left ( \frac{9}{5\sqrt{10}} \right )
 

Option 2)

\tan ^{-1}\left ( \frac{9}{5\sqrt{10}} \right )

Option 3)

\cos ^{-1}\left ( \frac{9}{5\sqrt{10}} \right )

 

Option 4)

\tan ^{-1}\left ( \frac{9}{14} \right )

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