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TS EAMCET · Maths · Inverse Trigonometric Functions

If \(\operatorname{sech}^{-1}\left(\frac{1}{2}\right)-\operatorname{cosec~h}^{-1}\left(\frac{3}{4}\right)=\log _e k\), then

  1. A \(3 k^2-12 k-1=0\)
  2. B \(3 k^2-12 k+1=0\)
  3. C \(9 k^2-12 k+1=0\)
  4. D \(9 k^2-12 k-1=0\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(9 k^2-12 k+1=0\)

Step-by-step Solution

Detailed explanation

Given, \(\operatorname{sech}^{-1}\left(\frac{1}{2}\right)-\operatorname{cosech}^{-1}\left(\frac{3}{4}\right)=\log _e k\) \(\Rightarrow \log _e\left(\sqrt{2^2-1}+2\right)-\log _e\left(\sqrt{\left(\frac{4}{3}\right)^2}+1+\frac{4}{3}\right)=\log _e k\)…