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Please solve RD Sharma class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 21 maths textbook solution

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Answer: 20^{\circ}

Given:    \cos ^{-1}\left(\cos 350^{\circ}\right)-\sin ^{-1}\left(\sin 350^{\circ}\right)

Hint: \left[\sin \left(360^{\circ}-x\right)=-\sin x, \cos \left(360^{\circ}-x\right)=\cos x\right]


            \cos ^{-1}\left(\cos 350^{\circ}\right)-\sin ^{1}\left(\sin 350^{\circ}\right) \\

       =\cos ^{-1}\left\{\cos \left(360^{\circ}-350^{\circ}\right)\right\}-\sin ^{-1}\left\{\sin \left(360^{\circ}-350^{\circ}\right)\right\} \\

      =\cos ^{-1}\left\{\cos \left(10^{\circ}\right)\right\}-\sin ^{-1}\left\{\sin \left(-10^{\circ}\right)\right\} \\\\

     =10^{\circ}-\left(-10^{\circ}\right) \\

     = 20 \degree

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