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  • Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.11 Question 1 Subquestion (i) Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.11 Question 1 Subquestion (i) Maths Textbook Solution

Answers (1)

Answer:
\tan ^{-1}\left ( \frac{2}{9} \right )
Hint:
There is one formula in trigonometric function for union.
Let’s see the formula,
\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left ( \frac{x+y}{1-xy} \right )
Given:
\tan^{-1}\left ( \frac{1}{7} \right )+\tan^{-1}\left ( \frac{1}{13} \right )=\tan^{-1}\left ( \frac{2}{9} \right )
In LHS side,
Let      x=\frac{1}{7} \: and\: y=\frac{1}{13}
Explanation:
Let’s take LHS
L.H.S  =\tan^{-1}\frac{1}{7}+\tan^{-1}\frac{1}{13}
Let’s use formula,
\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left ( \frac{x+y}{1-xy} \right )
Let’s put the value of x and y in formula
\begin{aligned} L.H.S &=\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{13} \\ &=\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{1}{13}}{1-\left(\frac{1}{7} \times \frac{1}{13}\right)}\right) \end{aligned}
              \begin{aligned} &=\tan ^{-1}\left(\frac{\frac{13+7}{91}}{\frac{91-1}{91}}\right) \\ &=\tan ^{-1}\left(\frac{13+7}{91-1}\right) \\ &=\tan ^{-1}\left(\frac{20}{90}\right) \\ &=\tan ^{-1}\left(\frac{2}{9}\right) \end{aligned}
              =R.H.S
Hence, the prove
Note: We must remember the formula of union.

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