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Name: Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 23 math
Uploaded: Fri, 2021-08-06 16:38
Description: Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 23 math

#CBSE 12 Class
#School
#Maths
#RD Sharma Vol. 1 Class 12
#R.D. Sharma
#Inverse Trigonometric Functions

#1.3
#3.1
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#3.3
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#3.11
#3.12
#9.12
#3.10
#Fill in the blanks
#Very Short Answer type
#Multiple Choice Questions (MCQs)
#18.1
#3.14
#3.13

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Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 23 math

Answers (1)

Answer: 1

Given:

$\tan ^{-1} x+\tan ^{-1} y=\frac{\pi}{4}$

Hint: Try to separate the $\tan$ function into RHS, So that the variables gets free.

Solution:

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

Now,

$\begin{array}{l} \tan ^{-1} x+\tan ^{-1} y=\frac{\pi}{4} \\\\ \tan ^{-1}\left(\frac{x+y}{1-x y}\right)=\frac{\pi}{4} \\\\ \frac{x+y}{1-x y}=\tan \frac{\pi}{4} \\\\ \frac{x+y}{1-x y}=1 \\\\ x+y=1-x y \\\\ x+y+x y=1 \end{array}$

Posted by

infoexpert22

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Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 23 math

Answers (1)

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