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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise 3.14 Question 2 Sub Question 10 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise 3.14 Question 2 Sub Question 10 Maths Textbook Solution.

Answers (1)

Answer: 4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}=\frac{\pi}{4}

Hints: First we will solve for 4 \tan ^{-1} \frac{1}{5}

Split into 2 \times 2 \tan ^{-1} \frac{1}{5} and solve it

Given: 4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}=\frac{\pi}{4}

Explanation: Let us first solve 4 \tan ^{-1} \frac{1}{5}

4 \tan ^{-1} \frac{1}{5}=2 \times 2 \tan ^{-1} \frac{1}{5}

=2 \tan ^{-1}\left(\frac{2 \times \frac{1}{5}}{1-\left(\frac{1}{5}\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{1}{5}<1 \end{array}\right]

\begin{aligned} 4 \tan ^{-1} \frac{1}{5} &=2 \tan ^{-1}\left(\frac{\frac{2}{5}}{1-\frac{1}{25}}\right) \\ &=2 \tan ^{-1}\left(\frac{2}{5} \times \frac{25}{24}\right) \\ &=2 \tan ^{-1}\left(\frac{5}{12}\right) \end{aligned}

=\tan ^{-1}\left(\frac{2 \times \frac{5}{12}}{1-\left(\frac{5}{12}\right)^{2}}\right) \quad\left[\begin{array}{l} \because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ -1<\frac{5}{12}<1 \end{array}\right]

\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{5}{6}}{1-\frac{25}{144}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{5}{6}}{\frac{119}{144}}\right) \\ &=\tan ^{-1}\left(\frac{5}{6} \times \frac{144}{119}\right) \end{aligned}

4 \tan ^{-1} \frac{1}{5}=\tan ^{-1}\left(\frac{120}{119}\right)        ......(1)

Now

L.H.S:

\begin{aligned} &4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239} \\ &=\tan ^{-1}\left(\frac{120}{119}\right)-\tan ^{-1}\left(\frac{1}{239}\right) \\ &=\tan ^{-1}\left(\frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119} \times \frac{1}{239}}\right) \end{aligned}                                \left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]

\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{28680-119}{28441}}{\frac{28441+120}{28441}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{28561}{28441}}{\frac{28561}{28441}}\right) \end{aligned}

\begin{aligned} &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}                                                                        \left[\begin{array}{l} \tan \frac{\pi}{4}=1 \\ \tan ^{-1}(1)=\frac{\pi}{4} \end{array}\right]

Hence it is proved that  4 \tan ^{-1} \frac{1}{5}-\tan ^{-1} \frac{1}{239}=\frac{\pi}{4}

Posted by

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