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  • Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 3 Maths Textbbok Solution.

Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 3 Maths Textbbok Solution.

Answers (1)

Answer: \tan ^{-1} \frac{2}{3}=\frac{1}{2} \tan ^{-1} \frac{12}{5}

Hint: First we will multiply in numerator and denominator by 2 in L.H.S so that we can use the formula of 2 \tan ^{-1} x

Given: \tan ^{-1} \frac{2}{3}=\frac{1}{2} \tan ^{-1} \frac{12}{5}

Explanation:

L.H.S: \tan ^{-1} \frac{2}{3}

On multiply in numerator and denominator by 2.

=\frac{1}{2} \times 2 \tan ^{-1} \frac{2}{3}

=\frac{1}{2} \times \tan ^{-1}\left(\frac{2 \times \frac{2}{3}}{1-\left(\frac{2}{3}\right)^{2}}\right)                            \left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1<x<1 \\ -1<\frac{2}{3}<1 \end{array}\right]

\begin{aligned} &=\frac{1}{2} \times \tan ^{-1}\left(\frac{\frac{4}{3}}{1-\frac{4}{9}}\right) \\ &=\frac{1}{2} \times \tan ^{-1}\left(\frac{\frac{4}{3}}{\frac{5}{9}}\right) \\ &=\frac{1}{2} \times \tan ^{-1}\left(\frac{4}{3} \times \frac{9}{5}\right) \\ &=\frac{1}{2} \tan ^{-1}\left(\frac{12}{5}\right) \end{aligned}

Hence it is proved.

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