Question

Which is greater, tan 1 or tan⁻¹ 1 ?

Answer 1

· 1 rad = 57.3 degree which is bigger than π / 4 = 45 degrees, the value where reaches the value of 1. Which means tan ⁡ 1 > 1 and.

Missing: ⁻ ‎| ‎Must include: ‎⁻

Answer 2

Answer:

We know,

${ \tan}^{ - 1} 1 = { \tan}^{ - 1} \bigg( \tan \dfrac{\pi}{4} \bigg) = \dfrac{\pi}{4}$

Here, we need to find which of the two given trigonometric ratios is greater. We will use the fact that the number π/4 is less than the number 1.

so,

$\implies \: 1 > \dfrac{\pi}{4}$

Now

$\implies \rm \: \tan1 > \tan \dfrac{\pi}{4} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

[tan theta is an increasing function]

$\implies \rm \tan1 > 1$

$\implies \rm \: \tan1 > 1 > \dfrac{ \pi}{4} \: \: \: \: \: \: \: \: \: \: \: \: \bigg[1 > \dfrac{\pi}{4} \bigg]$

$\implies \rm \tan1 > \dfrac{\pi}{4}$

$\implies \rm \: \tan1 > { \tan }^{ - 1} 1 \: \: \: \: \: \bigg[ { \tan }^{ - 1} = \dfrac{\pi}{4} \bigg]$

so, clearly tan 1 is greater than tan⁻¹ 1 . ( Ans)

Inverse Identities

$\longrightarrow \rm \sin( { \sin }^{ - 1} x) = x \: \: \: \: \: \: \: \: x\in [ - 1,1]$

$\longrightarrow \rm \cos ( { \cos }^{ - 1} x) = x \: \: \: \: \: \: \: \: x\in [ - 1,1]$

$\longrightarrow \rm \tan ( { \tan }^{ - 1} x) = x \: \: \: \: \: \: \: \: x\in R$

$\longrightarrow \rm \cot( { \cot }^{ - 1} x) = x \: \: \: \: \: \: \: \: x\in R$

$\longrightarrow \rm \sec ( { \sec }^{ - 1} x) = x \: \: \: \: x\in R - (- 1,1)$

$\longrightarrow \rm \csc( { \csc }^{ - 1} x) = x \: \: \: \: x\in R - (- 1,1)$

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