Get Answers to all your Questions

header-bg qa

Please solve RD Sharma class 12 chapter Inverse Trignometric Functions exercise 3.7 question 4 sub question (iii) maths textbook solution

Answers (1)


Answer: \frac{3\pi }{4}

Hint: The range of principal value of  \sec ^{-1} is \left [ 0,\pi \right ]-\left [ \frac{\pi }{2} \right ]
Given:  \sec ^{-1}\left(\sec \frac{5 \pi}{4}\right)


First we solve  \sec \frac{5 \pi}{4}

But \left(2 \pi-\frac{3 \pi}{4}\right)  belongs to 3 quadrant

            So, \sec \left(2 \pi-\frac{3 \pi}{4}\right)=\sec \frac{3 \pi}{4}

            \begin{aligned} &\therefore \sec \frac{3 \pi}{4}=\sqrt{2} \\ &-\sec \frac{3 \pi}{4}=-\sqrt{2} \end{aligned}

By substituting these value in  \sec ^{-1}\left(\sec \frac{5 \pi}{4}\right)  we get,

            \sec ^{-1}(-\sqrt{2})

Now,     \text { let } y=\sec ^{-1}(-\sqrt{2})

\Rightarrow \quad \sec y=-\sqrt{2}

\Rightarrow \quad-\sec \left(\frac{\pi}{4}\right)=\sqrt{2}

\begin{aligned} &\Rightarrow \quad \sec \left(\pi-\frac{\pi}{4}\right) \\ &\Rightarrow \quad \sec \left(\frac{3 \pi}{4}\right)=-\sqrt{2} \end{aligned}

The range of principal value of  \sec ^{-1} \text { is }[0, \pi]-\left\{\frac{\pi}{2}\right\}

            \sec ^{-1}\left(\sec \frac{3 \pi}{4}\right)=\frac{3 \pi}{4} \; \; \; \; \; \; \; \; \; \; \quad \therefore \sec ^{-1}(\sec x)=x, x \in[0, \pi]-\left\{\frac{\pi}{2}\right\}

Posted by


View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support