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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise 3.14 Question 11 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise 3.14 Question 11 Maths Textbook Solution.

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Answer: \frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)

Given: For any a, b, x, y > 0

            We have to prove that \frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right)

            where,\alpha=-a x+b y \& \beta=b x+a y

Hint:  First we will divide numerator and denominator of first function and second function by b^{3} and y^{3}respectively, then use \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)

Solution: Let us assume

              \begin{aligned} L . H . S &=\frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right) \\ &=\frac{2}{3} \tan ^{-1}\left[\frac{\frac{3 a b^{2}-a^{3}}{b^{3}}}{\frac{b^{3}-3 a^{2} b}{b^{3}}}\right]+\frac{2}{3} \tan ^{-1}\left[\frac{\frac{3 x y^{2}-x^{3}}{y^{3}}}{\frac{y^{3}-3 x^{2} y}{y^{3}}}\right] \end{aligned}

Dividing numerator and denominator of first function and second function by b^{3} and y^{3}respectively.

                           =\frac{2}{3} \tan ^{-1}\left[\frac{3\left(\frac{a}{b}\right)-\left(\frac{a}{b}\right)^{3}}{1-3\left(\frac{a}{b}\right)^{2}}\right]+\frac{2}{3} \tan ^{-1}\left[\frac{3\left(\frac{x}{y}\right)-\left(\frac{x}{y}\right)^{3}}{1-3\left(\frac{x}{y}\right)^{2}}\right]

We know that,3 \tan ^{-1} x=\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)

            \Rightarrow \mathrm{LH} . \mathrm{S}=\frac{2}{3}\left[3 \tan ^{-1}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)\right]+\frac{2}{3}\left[3 \tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)\right]

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

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

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

                               =\tan ^{-1}\left(\frac{2 \times \frac{\beta}{\alpha}}{1-\left(\frac{\beta}{\alpha}\right)^{2}}\right)                                                                               \left[2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right]

                                \begin{aligned} &=\tan ^{-1}\left(\frac{2 \beta}{\alpha} \times \frac{\alpha^{2}}{\alpha^{2}-\beta^{2}}\right) \\ &=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right) \\ &=\text { R.H.S } \end{aligned}

Hence,\frac{2}{3} \tan ^{-1}\left(\frac{3 a b^{2}-a^{3}}{b^{3}-3 a^{2} b}\right)+\frac{2}{3} \tan ^{-1}\left(\frac{3 x y^{2}-x^{3}}{y^{3}-3 x^{2} y}\right)=\tan ^{-1}\left(\frac{2 \alpha \beta}{\alpha^{2}-\beta^{2}}\right) Where \alpha=-a x+b y \& \beta=b x+a y   

                                                                                                                                                         

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