Solve the following:
(a) Evaluate $\sin \left( {{\cos }^{-1}}\dfrac{3}{5} \right)$
(b) Evaluate $\cos \left( {{\tan }^{-1}}\dfrac{3}{4} \right)$ and $\cos \left( {{\tan }^{-1}}x \right)$
(c) If $\sin \left( {{\cot }^{-1}}\left( 1+x \right) \right)=\cos \left( {{\tan }^{-1}}x \right)$ then $x$ is
    (a) $\dfrac{1}{2}$
    (b) $1$
    (c) $0$
    (d) $-\dfrac{1}{2}$
(d) Evaluate $\sin \left( {{\cot }^{-1}}x \right)$
(e) Evaluate $\sin \left( 2{{\sin }^{-1}}0.8 \right)$

Hint: We will assign an angle to the inverse trigonometric functions. Then we will rearrange the equations to simplify them. We will use the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ to substitute certain terms in the equations and find the requires values.

Complete step by step answer:
(a) $\sin \left( {{\cos }^{-1}}\dfrac{3}{5} \right)$
Let ${{\cos }^{-1}}\dfrac{3}{5}=\theta $. So we have $\cos \theta =\dfrac{3}{5}$.
Using the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we know that ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $. Substituting the value of $\cos \theta $ in this equation, we get ${{\sin }^{2}}\theta =1-{{\left( \dfrac{3}{5} \right)}^{2}}=1-\dfrac{9}{25}=\dfrac{16}{25}$ . Therefore, by taking square root on both sides of this equation, we get $\sin \theta =\dfrac{4}{5}$ . That is, $\sin \left( {{\cos }^{-1}}\dfrac{3}{5} \right)=\dfrac{4}{5}$.
(b) $\cos \left( {{\tan }^{-1}}\dfrac{3}{4} \right)$
Let ${{\tan }^{-1}}\dfrac{3}{4}=\theta $. Therefore, we have $\tan \theta =\dfrac{3}{4}$. We can write $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and we can substitute $\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }$. Therefore, we have the following equation,
$\dfrac{\sqrt{1-{{\cos }^{2}}\theta }}{\cos \theta }=\dfrac{3}{4}$ .
Squaring both sides and rearranging the equation, we get
$\begin{align}
  & \dfrac{\sqrt{1-{{\cos }^{2}}\theta }}{\cos \theta }=\dfrac{3}{4} \\
 & 16\times (1-{{\cos }^{2}}\theta )=9\times {{\cos }^{2}}\theta \\
 & 16-16{{\cos }^{2}}\theta =9{{\cos }^{2}}\theta \\
 & 25{{\cos }^{2}}\theta =16 \\
 & {{\cos }^{2}}\theta =\dfrac{16}{25} \\
 & \cos \theta =\dfrac{4}{5} \\
\end{align}$
Therefore, resubstituting the value of $\theta $, we get $\cos \left( {{\tan }^{-1}}\dfrac{3}{4} \right)=\dfrac{4}{5}$.
Now, we will use similar method to find the value of $\cos ({{\tan }^{-1}}x)$. Let ${{\tan }^{-1}}x=\theta $.
Therefore, we have $\tan \theta =x$. Writing $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and substituting $\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }$, we get
$\dfrac{\sqrt{1-{{\cos }^{2}}\theta }}{\cos \theta }=x$ .
Squaring both sides and rearranging the equation, we get
$\begin{align}
  & \dfrac{\sqrt{1-{{\cos }^{2}}\theta }}{\cos \theta }=x \\
 & (1-{{\cos }^{2}}\theta )={{x}^{2}}\times {{\cos }^{2}}\theta \\
 & {{\cos }^{2}}\theta +{{x}^{2}}{{\cos }^{2}}\theta =1 \\
 & (1+{{x}^{2}}){{\cos }^{2}}\theta =1 \\
 & {{\cos }^{2}}\theta =\dfrac{1}{1+{{x}^{2}}} \\
 & \cos \theta =\dfrac{1}{\sqrt{1+{{x}^{2}}}} \\
\end{align}$
Therefore, we have $\cos ({{\tan }^{-1}}x)=\dfrac{1}{\sqrt{1+{{x}^{2}}}}$ .
(c) $\sin \left( {{\cot }^{-1}}\left( 1+x \right) \right)=\cos \left( {{\tan }^{-1}}x \right)$
From the previous question, we know that the RHS is $\cos ({{\tan }^{-1}}x)=\dfrac{1}{\sqrt{1+{{x}^{2}}}}$. Let ${{\cot }^{-1}}(1+x)=\theta $.
Therefore, we have $\cot \theta =1+x$. Now, we will write $\cos \theta =\dfrac{\cos \theta }{\sin \theta }$ and substitute $\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }$. We get the following equation,
$\dfrac{\sqrt{1-{{\sin }^{2}}\theta }}{\sin \theta }=1+x$
Squaring both sides and rearranging, we get
$\begin{align}
  & \dfrac{1-{{\sin }^{2}}\theta }{{{\sin }^{2}}\theta }={{(1+x)}^{2}} \\
 & \dfrac{1}{{{\sin }^{2}}\theta }-1=1+2x+{{x}^{2}} \\
 & \dfrac{1}{{{\sin }^{2}}\theta }=2+2x+{{x}^{2}} \\
 & \sin \theta =\dfrac{1}{\sqrt{2+2x+{{x}^{2}}}} \\
\end{align}$
Therefore, the LHS is $\sin \left( {{\cot }^{-1}}(1+x) \right)=\dfrac{1}{\sqrt{2+2x+{{x}^{2}}}}$ . So, now equating LHS and RHS, we get
$\dfrac{1}{\sqrt{2+2x+{{x}^{2}}}}=\dfrac{1}{\sqrt{1+{{x}^{2}}}}$
Squaring both sides and rearranging, we get
$\begin{align}
  & \dfrac{1}{2+2x+{{x}^{2}}}=\dfrac{1}{1+{{x}^{2}}} \\
 & 1+{{x}^{2}}=2+2x+{{x}^{2}} \\
 & 1-2=2x \\
 & x=-\dfrac{1}{2} \\
\end{align}$
So, the correct option is (d).
(d) $\sin \left( {{\cot }^{-1}}x \right)$
Let ${{\cot }^{-1}}x=\theta $. We will write $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ and substitute $\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }$.
We will get the following,
$\begin{align}
  & \cot \theta =x \\
 & \dfrac{\cos \theta }{\sin \theta }=x \\
 & \dfrac{\sqrt{1-{{\sin }^{2}}\theta }}{\sin \theta }=x \\
\end{align}$
Squaring both sides and rearranging, we get
$\begin{align}
  & \dfrac{1-{{\sin }^{2}}\theta }{{{\sin }^{2}}\theta }={{x}^{2}} \\
 & \dfrac{1}{{{\sin }^{2}}\theta }-1={{x}^{2}} \\
 & \dfrac{1}{{{\sin }^{2}}\theta }={{x}^{2}}+1 \\
 & {{\sin }^{2}}\theta =\dfrac{1}{{{x}^{2}}+1} \\
 & \sin \theta =\dfrac{1}{\sqrt{{{x}^{2}}+1}} \\
\end{align}$
Therefore, we have $\sin \left( {{\cot }^{-1}}x \right)=\dfrac{1}{\sqrt{1+{{x}^{2}}}}$.
(e) $\sin \left( 2{{\sin }^{-1}}0.8 \right)$
Let ${{\sin }^{-1}}0.8=\theta $. Therefore, we have $\sin \theta =0.8$. Using the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get
$\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }=\sqrt{1-{{(0.8)}^{2}}}=\sqrt{1-0.64}=\sqrt{0.36}=0.6$.
Now, $\sin \left( 2{{\sin }^{-1}}0.8 \right)=\sin (2\theta )$ and we have the identity $\sin 2\theta =2\sin \theta \cos \theta $. Substituting the values of $\sin \theta $ and $\cos \theta $, we get
$\sin \left( 2{{\sin }^{-1}}0.8 \right)=2\times 0.8\times 0.6=0.96$

Note:
To solve this question, we used trigonometric identities and relations. It is important to remember these as these are very useful. Solving such questions becomes easier in this manner. We should be careful while substituting and resubstituting values, as there is a possibility to go around in a loop while trying to simplify trigonometric equations.