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Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise Multiple Choice Questions Question 4 Maths Textbook Solution.

Answers (1)

Answer: \sin ^{2} \alpha

Hint: Let \cos \alpha get into RHS, so that variable can free


        If,\cos ^{-1} \frac{x}{a}+\cos ^{-1} \frac{y}{b}=\alpha

        \frac{x^{2}}{a^{2}}-\frac{2 x y}{a b} \cos \alpha+\frac{y^{2}}{b^{2}}=?


We know that,

            \begin{aligned} &\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right. \\ &\therefore \cos ^{-1}\left(\frac{x}{a}\right)+\cos ^{-1}\left(\frac{y}{b}\right)=\alpha \end{aligned}

            \begin{aligned} &\cos ^{-1}\left(\frac{x}{a} \frac{y}{b}-\sqrt{1-\frac{x^{2}}{a^{2}}} \sqrt{1-\frac{y^{2}}{b^{2}}}\right)=\alpha \\ &\frac{x y}{a b}-\sqrt{1-\frac{x^{2}}{a^{2}}} \sqrt{1-\frac{y^{2}}{b^{2}}}=\cos \alpha \\ &\sqrt{1-\frac{x^{2}}{a^{2}}} \sqrt{1-\frac{y^{2}}{b^{2}}}=\frac{x y}{a b}-\cos \alpha \end{aligned}

Squaring on both the sides,

          \left(1-\frac{x^{2}}{a^{2}}\right)\left(1-\frac{y^{2}}{b^{2}}\right)=\frac{x^{2}}{y^{2}} \frac{y^{2}}{b^{2}}+\cos ^{2} \alpha-\frac{2 x y}{a b} \cos \alpha        Mention a square instead of y square in denominator

           \begin{aligned} &1-\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}+\frac{x^{2}}{a^{2}} \frac{y^{2}}{b^{2}}=\frac{x^{2}}{a^{2}} \frac{y^{2}}{b^{2}} \\ &\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{2 x y}{a b} \cos \alpha=1-\cos ^{2} \alpha \end{aligned}

                                                     =\sin ^{2} \alpha                  MEntion how the elimination of last two term has been done

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