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  • Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.11 Question 3 Subquestion (xi) Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.11 Question 3 Subquestion (xi) Maths Textbook Solution.

Answers (1)

Answer:
x= \left ( \frac{1}{12} \right ),\left ( \frac{-1}{2} \right )
Hint:
Here, we use the below formula
\tan^{-1}A+\tan^{-1}B=\tan^{-1}\left ( \frac{A+B}{1-AB} \right )
Given:
\tan^{-1}4x+\tan^{-1}6x=\frac{\pi }{4}
Solution:
Here      A=4x
             B=6x
Let’s put the values of A and B in formula of\tan^{-1}A+\tan^{-1}B ,we get
\tan^{-1}\left ( \frac{4x+6x}{1-(4x)(6x)} \right )=\frac{\pi }{4}
\Rightarrow \tan^{-1}\left ( \frac{10x}{1-24x^{2}} \right )=\frac{\pi }{4}
\Rightarrow \frac{10x}{1-24x^{2}}=\tan \frac{\pi }{4}
\Rightarrow \frac{10x}{1-24x^{2}}=1                                                                                        \left ( \tan \left ( \frac{\pi }{4} \right )= 1 \right )
\! \! \! \! \! \! \Rightarrow 10x=1-24x^{2}\\ \Rightarrow 24x^{2}+10x-1=0\\ \Rightarrow 24x^{2}+12x-2x-1=0\\ \Rightarrow 12x\left ( 2x+1 \right )-1\left ( 2x+1 \right )=0\\ \Rightarrow \left ( 12x-1 \right )\left ( 2x+1 \right )=0\\
\! \! \! \! \! \! \! \! \! \Rightarrow Either\: 12x-1=0 \: or \: 2x+1=0\\ \Rightarrow 12x=1 \: or \: 2x= -1 \\ \Rightarrow x=\frac{1}{12} \: or \: x =-\frac{1}{2}\\ \Rightarrow x=\frac{1}{12},\left ( \frac{-1}{2} \right )\\

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