Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise 3.14 Question 3 Maths Textbook Solution.

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

Answers (1)

Answer: \sin ^{-1} \frac{2 a}{1+a^{2}}-\cos \frac{1-b^{2}}{1+b^{2}}=\tan ^{-1} \frac{2 x}{1-x^{2}} \text { then } x=\frac{a-b}{1+a b}

Hints: First we will convert whole L.H.S part in \tan ^{-1}

Given:\sin ^{-1} \frac{2 a}{1+a^{2}}-\cos \frac{1-b^{2}}{1+b^{2}}=\tan ^{-1} \frac{2 x}{1-x^{2}}

Explanation:

As we know that

\begin{aligned} &2 \tan ^{-1} x=\sin ^{-1} \frac{2 x}{1+x^{2}}, 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right) \\ &2 \tan ^{1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \end{aligned}

Now \sin ^{-1} \frac{2 a}{1+a^{2}}-\cos \frac{1-b^{2}}{1+b^{2}}=\tan ^{-1} \frac{2 x}{1-x^{2}}

\begin{aligned} &2 \tan ^{-1} a-2 \tan ^{-1} b=\tan ^{-1} \frac{2 x}{1-x^{2}} \\ &2\left(\tan ^{-1} a-\tan ^{-1} b\right)=\tan ^{-1} \frac{2 x}{1-x^{2}} \\ &2 \tan ^{-1}\left(\frac{a-b}{1+a b}\right)=\tan ^{-1} \frac{2 x}{1-x^{2}} \\ &2 \tan ^{-1}\left(\frac{a-b}{1+a b}\right)=2 \tan ^{-1} x \\ &\tan ^{-1}\left(\frac{a-b}{1+a b}\right)=\tan ^{-1} x \\ &\frac{a-b}{1+a b}=\tan \left(\tan ^{-1} x\right) \\ &\frac{a-b}{1+a b}=x \\ &x=\frac{a-b}{1+a b} \end{aligned}

Hence it is proved that x=\frac{a-b}{1+a b}

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

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
anam-khan

When will CBSE Board exam 2025 date sheet be released for Class 10, 12?

Latest Question