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Name: Explain solution for RD Sharma class class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 16 math
Uploaded: Thu, 2021-08-05 23:21
Description: Explain solution for RD Sharma class class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 16 math

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

#1.3
#3.1
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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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Explain solution for RD Sharma class class 12 chapter 3 Inverse Trigonometric Functions exercise Very short answer question 16 math

Answers (1)

Answer: $70 \degree$

Given:

$\sin ^{-1}\left(\sin 1550^{\circ}\right)$

Hint:

$\sin ^{-1}(\sin x)=x$

Solution:

We, know that

$\sin ^{-1}(\sin x)=x$

Now,

$\begin{aligned} \sin ^{-1}\left(\sin 1550^{\circ}\right) &=\sin ^{-1}\left\{\sin \left(1620^{\circ}-1550^{\circ}\right)\right\} \\ &=\sin ^{-1}\left(\sin 70^{\circ}\right) \\ &=70^{\circ} \end{aligned}$

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 16 math

Answers (1)

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