Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 7 Maths Textbook Solution.

Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise 3.14 Question 2 sub question 7  Maths Textbook Solution.

Answers (1)

Answer:2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\tan ^{-1} \frac{4}{7}

Hint: First we will solve for 2 \tan ^{-1} \frac{1}{5}

Given: 2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\tan ^{-1} \frac{4}{7}

Explanation:

L.H.S:

2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}                                            

=\tan ^{-1}\left(\frac{\frac{2}{5}}{1-\left(\frac{1}{5}\right)^{2}}\right)+\tan ^{-1}\left(\frac{1}{8}\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} &=\tan ^{-1}\left(\frac{\frac{2}{5}}{1-\left(\frac{1}{25}\right)}\right)+\tan ^{-1}\left(\frac{1}{8}\right) \\ &=\tan ^{-1}\left(\frac{\frac{2}{5}}{\frac{24}{25}}\right)+\tan ^{-1}\left(\frac{1}{8}\right) \\ &=\tan ^{-1}\left(\frac{2}{5} \times \frac{25}{24}\right)+\tan ^{-1}\left(\frac{1}{8}\right) \\ &=\tan ^{-1}\left(\frac{5}{12}\right)+\tan ^{-1}\left(\frac{1}{8}\right) \end{aligned}

                                                                                                            \left[\begin{array}{l} \because \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \\ x y<1 \\ \frac{5}{12} \times \frac{1}{8}=\frac{5}{96}<1 \end{array}\right]

\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{5}{12}+\frac{1}{8}}{1-\frac{5}{12} \times \frac{1}{8}}\right) \\ &=\tan ^{-1}\left(\frac{\frac{10+3}{24}}{1-\frac{5}{96}}\right) \end{aligned}

\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{13}{24}}{\frac{91}{96}}\right) \\ &=\tan ^{-1}\left(\frac{13}{24} \times \frac{96}{91}\right) \\ &=\tan ^{-1}\left(\frac{4}{7}\right) \end{aligned}

Hence it is proved that 2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\tan ^{-1} \frac{4}{7}

Posted by

infoexpert21

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

Board Exam Date Sheet 2025: CBSE, UPMSP, MPBSE, Class 10, 12 board exam dates out; BSEB, CGBSE schedules soon
anam-khan

CBSE Date Sheet 2025 (Out) Live: Class 10, 12 board exam time table at cbse.gov.in; syllabus, sample papers
anam-khan

CBSE Class 12 board exam date sheet 2025 out; exams from February 15 to April 4

Latest Question