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 RD Sharma maths class 12 chapter 10 Differentiation exercise Very short answers question 15

Need solution for RD Sharma maths class 12 chapter 10 Differentiation exercise Very short answers question 15

Answers (1)

Answer:

    -1

Hint:

\begin{aligned} &\text { Using (1) } 1-\cos 2 x=2 \sin ^{2} x\\ &\text { (2) } 1+\cos 2 x=2 \cos ^{2} x \end{aligned}

Given:

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

Solution:  

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

Using

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

\begin{aligned} &y=\tan ^{-1}(-\tan x)\left[-\frac{\pi}{2}<x<0\right] \\\\ &y=-x \quad\left[\because \tan ^{-1}(\tan x)=x\right] \end{aligned}

Differentiate it w.r.t x, we get

\begin{aligned} &\frac{d y}{d x}=\frac{d(-x)}{d x} \\\\ &\frac{d y}{d x}=-1 \text { for } x \in\left(-\frac{\pi}{2}, 0\right) \end{aligned}

Posted by

infoexpert26

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 Central Sector Scheme of Scholarship 2025 applications open on scholarships.gov.in
anam-khan

CBSE Board 2025: Class 10th, 12th supplementary date sheet out; exams from July 15
anam-khan

CBSE issues final reminder to schools for supplementary exam LOC submission; submit by today without late fee

Latest Question