Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain Solution R D Sharma Class 12 Chapter 3 Inverse Trigonometric Function Exercise 3 point 14 Question 8 Sub Question 6 maths Textbook Solution

Explain Solution R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Function Exercise 3.14 Question 8 Sub Question 6 maths Textbook Solution.

Answers (2)

Answer: x=\sqrt{3}

Given:\cos ^{-1}\left(\frac{x^{2}-1}{x^{2}+1}\right)+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}

Hint: Using formula

         \text { 1. } \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \tan ^{-1} x

         \text { 2. } \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x

Solution:

             We have,

             \cos ^{-1}\left(\frac{x^{2}-1}{x^{2}+1}\right)+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3}

We know that,

            \begin{aligned} &\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=2 \tan ^{-1} x \\ &\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x \\ &\Rightarrow \cos ^{-1}\left[\frac{-\left(1-x^{2}\right)}{1+x^{2}}\right]+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3} \end{aligned}                                                [\cos (-\theta)=\pi-\cos \theta]

            \begin{aligned} &\Rightarrow \pi-\cos ^{-1}\left[\frac{\left(1-x^{2}\right)}{1+x^{2}}\right]+\frac{1}{2} \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{2 \pi}{3} \\ &\Rightarrow \pi-2 \tan ^{-1}(x)+\frac{1}{2} \times 2 \tan ^{-1} x=\frac{2 \pi}{3} \end{aligned}

 \left[\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=2 \tan ^{-1} x\right]

                  \begin{aligned} &\Rightarrow \pi-2 \tan ^{-1}(x)+\tan ^{-1}(x)=\frac{2 \pi}{3} \\ &\Rightarrow \pi-\tan ^{-1} x=\frac{2 \pi}{3} \\ &\Rightarrow \tan ^{-1} x=\pi-\frac{2 \pi}{3} \\ &\Rightarrow \tan ^{-1} x=\frac{\pi}{3} \\ &\Rightarrow x=\tan \frac{\pi}{3} \\ &\Rightarrow x=\sqrt{3} \end{aligned}                             \left[\tan \left(\frac{x}{3}\right)=\sqrt{3}\right]

Hence x=\sqrt{3} is required solution.

 

 

Posted by

infoexpert21

View full answer

Crack CUET with india's "Best Teachers"

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

Answer: x=\pm \sqrt{\frac{7}{2}}

Given: \tan ^{-1}\left(\frac{x-2}{x-1}\right)+\tan ^{-1}\left(\frac{x+2}{x+1}\right)=\frac{\pi}{4}

Hint: Use \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)

Solution: We have,

              \tan ^{-1}\left(\frac{x-2}{x-1}\right)+\tan ^{-1}\left(\frac{x+2}{x+1}\right)=\frac{\pi}{4}

We know that,

            \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)

            \begin{aligned} &\Rightarrow \tan ^{-1}\left[\frac{\frac{x-2}{x-1}+\frac{x+2}{x+1}}{1-\frac{x-2}{x-1} \times \frac{x+2}{x+1}}\right]=\frac{\pi}{4} \\ &\Rightarrow \tan ^{-1}\left[\frac{\frac{(x-2)(x+1)+(x-1)(x+2)}{\frac{x^{2}-1-[(x-2)(x+2)]}{x^{2}-1}}}{\frac{x^{2}-1}{4}}=\frac{\pi}{4}\right. \\ &\Rightarrow \tan ^{-1}\left[\frac{x^{2}+x-2 x-2+x^{2}+2 x-x-2}{x^{2}-1-x^{2}+4}\right]=\frac{\pi}{4} \end{aligned}

            \Rightarrow \tan ^{-1}\left[\frac{2 x^{2}-4}{3}\right]=\tan ^{-1}(1)                                                                            \left[\tan ^{-1}(1)=\frac{\pi}{4}\right]

            \begin{aligned} &\Rightarrow \frac{2 x^{2}-4}{3}=1 \\ &\Rightarrow 2 x^{2}=3+4 \\ &\Rightarrow 2 x^{2}=7 \end{aligned}

           \begin{aligned} &\Rightarrow x^{2}=\frac{7}{2} \\ &\Rightarrow x=\pm \sqrt{\frac{7}{2}} \end{aligned}

          Hence,  x=\pm \sqrt{\frac{7}{2}} is required solution

Posted by

infoexpert21

View full answer

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