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Name: Explain solution for RD Sharma class class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 36 math
Uploaded: Fri, 2021-08-06 22:16
Description: Explain solution for RD Sharma class class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 36 math

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

#1.3
#3.1
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#3.9
#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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Explain solution for RD Sharma class class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 36 math

Answers (1)

Answer: $\frac{1}{2}$

Given:

$4 \sin ^{-1} x+\cos ^{-1} x=\pi$

Hint:

$\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$

Solution:

$\begin{array}{l} \therefore 4 \sin ^{-1} x+\cos ^{-1} x=\pi \\\\ 4 \sin ^{-1} x+\frac{\pi}{2}-\sin ^{-1} x=\pi \quad\left[\therefore \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right] \\\\ 3 \sin ^{-1} x=\frac{\pi}{2} \\\\ \sin ^{-1} x=\frac{\pi}{6} \\\\ x=\sin \frac{\pi}{6} \\\\ x=\frac{1}{2} \end{array}$

Posted by

infoexpert22

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Explain solution for RD Sharma class class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 36 math

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