Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.11 Question 2 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.11 Question 2 Maths Textbook Solution.

Answers (1)

Answer:
\frac{\pi }{4}
Hint:
For solving this, we can use the formula of union trigonometric function,
\tan^{-1}a-\tan^{-1}b=\tan^{-1}\left ( \frac{a-b}{1+ab} \right )
Given:
\tan^{-1}\left ( \frac{x}{y} \right )-\tan^{-1}\left ( \frac{x-y}{x+y} \right )
Explanation:
Let’s use the formula
Wherea=\frac{x}{y}\: and \: b=\frac{x-y}{x+y}
\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right)=\tan ^{-1}\left(\frac{\frac{x}{y}-\left(\frac{x-y}{x+y}\right)}{1+\left(\frac{x}{y}\right)\left(\frac{x-y}{x+y}\right)}\right)
\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{x^{2}+x y-y(x-y)}{y(x+y)}}{\frac{x y+y^{2}+x^{2}-x y}{y(x+y)}}\right) \\ &=\tan ^{-1}\left(\frac{x^{2}+y^{2}}{x^{2}+y^{2}}\right) \\ &=\tan ^{-1}(1) \\ &=\frac{\pi}{4} \end{aligned}
Note: We must know the formula of union trigonometric function.




                                                

Posted by

infoexpert27

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