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 \(\tan(\sin^{-1}(\frac{x}{2}))=\frac{x}{\sqrt{4-x^2}}\)
