Get Answers to all your Questions

header-bg qa

Need solution for RD Sharma maths class 12 Chapter Inverse Trigonometric functions exercise 3.8 question 1 sub question (viii)

Answers (1)




As we know that the range of \cot \left(\cot ^{-1} x\right) \text { is } x \text { for } x \in \mathbb{R} \text { ( real number) }


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


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

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

Let’s suppose that,

\begin{aligned} &\cos ^{-1} \left (\frac{3}{5} \right )=\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 + (BC)^2 \\ & (BC)^2= 16\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} &\cot \left(\cos ^{-1}\left(\frac{3}{5}\right)\right)=\cot c \\ &\cot \alpha=\frac{3}{4} \\ &\cot \left(\cos ^{-1}\left(\frac{3}{5}\right)\right)=\frac{3}{4} \end{aligned}


Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support