Question

1. What’s the exact value of cot[cot^-1 sqrt3/3] ?

Answer 1

So the problems ask to find and calculate the exact value of the trigonometric equation in the following equations and the best answers would be the following:
#1. sqrt(3)/3
#2. Arcsine of zero is 0
#3. x/sqrt(4-x^2)
I hope you are satisfied with my answer and feel free to ask for more if you have questions and further clarifications. Have a nice day 

Answer 2

Answer and Explanation :

1) Expression $\cot[\cot^{-1}(\frac{\sqrt{3}}{3})]$

We have to find the exact solution of the expression.

We know the inverse property,

$\cot[\cot^{-1}x]=x$

Applying the property,

$\cot[\cot^{-1}(\frac{\sqrt{3}}{3})]=\frac{\sqrt{3}}{3}$    

Therefore, The exact solution is $\cot[\cot^{-1}(\frac{\sqrt{3}}{3})]=\frac{\sqrt{3}}{3}$        

2) Expression $\sin^{-1][\cos(\frac{\pi}{2})]$

We have to find the exact solution of the expression.

We know that,

$\cos(\frac{\pi}{2})=0$

$\sin(0)=0$

Applying the property,

$\sin^{-1][\cos(\frac{\pi}{2})]=\sin^{-1}[0]$

$\sin^{-1][\cos(\frac{\pi}{2})]=\sin^{-1}[\sin(0)]$

Applying inverse property,

$\sin^{-1}[\sin x]=x$

$\sin^{-1][\cos(\frac{\pi}{2})]=0$

Therefore, The exact solution is $\sin^{-1][\cos(\frac{\pi}{2})]=0$

3) Expression $\tan(\sin^{-1}(\frac{x}{2}))$

We have to find the exact solution of the expression.

Let   $\sin^{-1}(\frac{x}{2})=y$

i.e.    $\sin y=\frac{x}{2}$

Squaring both side,  $\sin^2 y=\frac{x^2}{4}$

Now, The expression became $\tan(y)$

We can write tan in form of sin and cosine as

$\tan y=\frac{\sin y}{\cos y}$

We know, $\cos y=\sqrt{1-\sin^2 y}$

Substituting,

$\tan y=\frac{\sin y}{\sqrt{1-\sin^2 y}}$

Now, put the value of sin y

$\tan y=\frac{\frac{x}{2}}{\sqrt{1-\frac{x^2}{4}}}$

$\tan y=\frac{\frac{x}{2}}{\sqrt{\frac{4-x^2}{4}}}$

$\tan y=\frac{\frac{x}{2}}{\frac{\sqrt{4-x^2}}{2}}$

$\tan y=\frac{x}{\sqrt{4-x^2}}$

Therefore, The exact solution is $\tan(\sin^{-1}(\frac{x}{2}))=\frac{x}{\sqrt{4-x^2}}$