Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please solve RD Sharma Chapter Inverse trigonometric functions exercise 3 8 question 1 sub question v maths text book solution

Please solve RD Sharma Chapter Inverse trigonometric functions exercise 3.8 question 1 sub question (v) maths text book solution

Answers (1)

\frac{5}{4}

Hint:

As we know that the value of \cos e c\left(\cos e c^{-1} x\right) \text { is } x \text { for } x \in(-\infty,-1) \cup(1, \infty)

Given:

We have

                \operatorname{cosec}\left(\cos ^{-1} \frac{3}{5}\right)

Solution:

In place of \cos e c^{-1} x here is \cos ^{-1} x

So, we convert \cos ^{-1} x into \cos e c^{-1} x.

Let’s suppose that,

\begin{aligned} &\cos ^{-1} \frac{3}{5}=\alpha \\ &\cos \alpha=\frac{3}{5}=\frac{B}{H} \end{aligned}

Let, in \Delta ABC,  angle CAB = \alpha and right angle at B.

\begin{aligned} &(A C)^{2}=(A B)^{2}+(B C)^{2} \\ &5^{2}=3^{2}+(B C)^{2} \\ &25-9=(B C)^{2} \end{aligned}

BC =\pm 4                    [we will ignore the -ve sign because BC is a length and it can’t be -ve]

BC =4

\begin{aligned} &\cos e c \alpha=\frac{H}{P}=\frac{5}{4} \\ &\cos e c\left(\cos ^{-1} \frac{3}{5}\right)=\frac{5}{4} \end{aligned}

 

 

Posted by

infoexpert24

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years

Latest Question