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Name: Provide solution for RD Sharma math class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 58
Uploaded: Sat, 2021-08-07 16:05
Description: Provide solution for RD Sharma math class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 58

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

#1.3
#3.1
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#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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Provide solution for RD Sharma math class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 58

Answers (1)

Answer: $\frac{2}{5}$

Given:

$\cos \left(\sin ^{-1} \frac{2}{5}+\cos ^{-1} x\right)=0$

Hint: Try to convert the separate $\sin$ , $\cos$ function. So, that the variable get free.

Solution:

$\begin{array}{l} \cos \left(\sin ^{-1} \frac{2}{5}+\cos ^{-1} x\right)=0 \\\\ \cos \left(\sin ^{-1} \frac{2}{5}+\cos ^{-1} x\right)=\cos \left(\frac{\pi}{2}\right) \\\\ \sin ^{-1} \frac{2}{5}+\cos ^{-1} x=\frac{\pi}{2} \\\\ x=\frac{2}{5} c g f y \end{array}$

Posted by

infoexpert22

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Provide solution for RD Sharma math class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 58

Answers (1)

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