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Name: Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 31 math
Uploaded: Fri, 2021-08-06 21:36
Description: Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 31 math

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

#1.3
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#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 31 math

Answers (1)

Answer: $2 \sin ^{-1} x=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)$

Given: To prove that

$\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x \\$

Hint: $x=\sin a$

Solution:

$\mathrm{LHS}=\sin ^{-1} 2 x \sqrt{1-x^{2}}$ Putting

$\begin {array} {ll} x=\sin a, we get \\\\ =\sin ^{-1}\left(2 \sin a \sqrt{1-\sin ^{2} a}\right) \\\\ =\sin ^{-1}(2 \sin a \cos a)\\\\ =\sin ^{-1}(\sin 2 a)\\\\ =2 a\\\\ =2 \sin ^{-1} x \quad[x=\sin a]\\\\ \end{array}$

$\therefore \sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right)=2 \sin ^{-1} x$

Posted by

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Need solution for RD Sharma maths class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 31 math

Answers (1)

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