TS EAMCET · Maths · Inverse Trigonometric Functions
\(\tanh ^{-1}\left(\frac{1}{2}\right)+\operatorname{coth}^{-1}(2)\) is equal to
- A \(\frac{1}{2} \log 3\)
- B \(\frac{1}{2} \log 6\)
- C \(\frac{1}{2} \log 12\)
- D \(\log 3\)
Answer & Solution
Correct Answer
(D) \(\log 3\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \tanh ^{-1}\left(\frac{1}{2}\right)+\operatorname{coth}^{-1}(2) \\ & \Rightarrow \tanh ^{-1}\left(\frac{1}{2}\right)+\tanh ^{-1}\left(\frac{1}{2}\right) \\ & \Rightarrow 2 \tanh ^{-1}\left(\frac{1}{2}\right)…
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