TS EAMCET · Maths · Inverse Trigonometric Functions
Consider the following statements
Assertion (A): For \(x \in \mathbb{R}-\{1\}, \frac{d}{d x}\left(\operatorname{Tan}^{-1}\left(\frac{1+x}{1-x}\right)\right)=\frac{d}{d x}\left(\operatorname{Tan}^{-1} x\right)\)
Reason (R): For \(x < 1, \operatorname{Tan}^{-1}\left(\frac{1+x}{1-x}\right)=\frac{\pi}{4}+\operatorname{Tan}^{-1} x\),
for \(x>1, \operatorname{Tan}^{-1}\left(\frac{1+x}{1-x}\right)=-\frac{3 \pi}{4}+\operatorname{Tan}^{-1} x\)
The correct answer is
- A Both (A) and (R) are true, (R) is the correct explanation of (A)
- B Both (A) and (R) are true, (R) is not the correct explanation of (A)
- C (A) is true, but (R) is false
- D (A) is false, but (R) is true
Answer & Solution
Correct Answer
(A) Both (A) and (R) are true, (R) is the correct explanation of (A)
Step-by-step Solution
Detailed explanation
\(\frac{d}{d x}\left(\operatorname{Tan}^{-1}\left(\frac{1+x}{1-x}\right)\right) = \frac{\frac{d}{d x}\left(\frac{1+x}{1-x}\right)}{1+\left(\frac{1+x}{1-x}\right)^2}\)…
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