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Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise 3.14 Question 10 Maths Textbook Solution.

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Answer: \tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)+\tan ^{-1}\left(\frac{2 x y}{x^{2}-y^{2}}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)

Given: \tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)+\tan ^{-1}\left(\frac{2 x y}{x^{2}-y^{2}}\right)where \alpha=a x-b y \: \: \& \: \: \beta=a y+b x

Hint: Use \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)

Solution: Let us assume

             L.H.S:=\tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)+\tan ^{-1}\left(\frac{2 x y}{x^{2}-y^{2}}\right)

             We know that \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)

                               \begin{aligned} &=\tan ^{-1}\left[\frac{\frac{2 a b}{a^{2}-b^{2}}+\frac{2 x y}{x^{2}-y^{2}}}{1-\left(\frac{2 a b}{a^{2}-b^{2}}\right)\left(\frac{2 x y}{x^{2}-y^{2}}\right)}\right] \\ &=\tan ^{-1}\left[\frac{2 a b x^{2}-2 a b y^{2}+2 x y a^{2}-2 x y b^{2}}{a^{2} x^{2}-a^{2} y^{2}-b^{2} x^{2}+b^{2} y^{2}-4 a b x y}\right] \\ &=\tan ^{-1}\left[\frac{2\left(a b x^{2}+x y a^{2}-a b y^{2}-x y b^{2}\right)}{a^{2} x^{2}-a^{2} y^{2}-b^{2} x^{2}+b^{2} y^{2}-4 a b x y}\right] \\ &=\tan ^{-1}\left[\frac{2[a x(b x+a y)-b y(a y+b x)]}{(a x-b y)^{2}-\left(a^{2} y^{2}+b^{2} x^{2}+2 a b x y\right)}\right] \end{aligned}

                               =\tan ^{-1}\left[\frac{2(b x+a y)(a x-b y)}{(a x-b y)^{2}-(b x+a y)^{2}}\right]                                                        \left[(a+b)^{2}=a^{2}+2 a b+b^{2}\right]

                               =\tan ^{-1}\left[\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right]                                                                                        [\alpha=a x-b y \& \beta=a y+b x]


Hence ,\tan ^{-1}\left(\frac{2 a b}{a^{2}-b^{2}}\right)+\tan ^{-1}\left(\frac{2 x y}{x^{2}-y^{2}}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)

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