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
- A \(3 k^2-12 k-1=0\)
- B \(3 k^2-12 k+1=0\)
- C \(9 k^2-12 k+1=0\)
- D \(9 k^2-12 k-1=0\)
Answer & Solution
Correct Answer
(C) \(9 k^2-12 k+1=0\)
Step-by-step Solution
Detailed explanation
Given, \(\operatorname{sech}^{-1}\left(\frac{1}{2}\right)-\operatorname{cosech}^{-1}\left(\frac{3}{4}\right)=\log _e k\) \(\Rightarrow \log _e\left(\sqrt{2^2-1}+2\right)-\log _e\left(\sqrt{\left(\frac{4}{3}\right)^2}+1+\frac{4}{3}\right)=\log _e k\)…
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