inverse trigonometric function

Sir / Madam,

In the following problem:

solve:

Cos(tan^{-1}x) = Sin[cot^{-1}(3/4)]

In the above said problem, I proceeded

sin[(Pi/2) - tan^{-1}x] = Sin[cot^{-1}(3/4)]

(Pi/2) - tan^{-1}x = cot^{-1}(3/4)

Cot^{-1}x = cot^{-1}(3/4)

x = 3/4

However, they have given x = + 3/4 or - 3/4

How to identify it has two answers?

How to find the second answer.

Kindly enlighten me.

with warm regards,

Aranga

Re: inverse trigonometric function

$\sin[\cot^{-1}(3/4)] = 4/5$

$\cos(\tan^{-1}{x}) = \dfrac{1}{\sqrt{1+x^2}}$

$\dfrac{1}{\sqrt{1+x^2}} = \dfrac{4}{5} \implies 1+x^2 = \dfrac{25}{16} \implies x = \pm \dfrac{3}{4}$

Re: inverse trigonometric function

As the cosine functiion is even and the inverse tangent function is odd, we may state:

$\displaystyle \cos\left(\arctan(x)\right)=\cos\left(\arctan(\pm x)\right)$