Q: Indicate the relation which is false
(A) tan | tan-1 x | = | x |
(B) cot | cot-1 x | = x
(C) tan-1 | tan x | = | x |
(D) sin | sin-1 x | = | x |
Sol. Since $ \large | tan^{-1}x | = \left\{\begin{array}{ll} tan^{-1}x \; , if \; 0 \leq tan^{-1}x \le \pi/2 \\ -tan^{-1}x \; , if \; -\pi/2 < tan^{-1}x < 0 \end{array} \right. $
$ \large = \left\{\begin{array}{ll} tan^{-1}x \; , if \; x \ge 0 \\ -tan^{-1}x \; , if x < 0 \end{array} \right. $
⇒ | tan-1x | = tan-1 |x| ∀ x ∈ R
⇒ tan| tan-1x | = tan tan-1 |x| = |x|
Hence (A) is the correct answer.
Likewise sin | sin-1x| = sin sin-1|x| = |x| ∀ |x| ≤ 1
Hence (D) is the correct answer.
| cot-1x | = cot-1x as 0 < | cot-1x| < π ∀ x ∈ R
⇒ cot |cot-1x| = cot cot-1x = x .
Hence (B) is the correct answer.
Since | tanx | is not necessarily always equal to tan|x|
Hence tan-1 | tanx | ≠ | x| .