Question

Prove the following trigonometric expression:
${{\sin }^{-1}}\left( \dfrac{5}{13} \right)+{{\cos }^{-1}}\left( \dfrac{3}{5} \right)={{\tan }^{-1}}\left( \dfrac{63}{16} \right)$

Answer 1

Hint: We will start with the L.H.S. of the equation we will assume both the terms as a and b then we will find out the value of$\tan a\text{ and }\tan b$, once we find out these value then we will apply the trigonometric property: $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$, from this we will get the value of $\left( a+b \right)$ in the form of $\arctan \text{ or }{{\tan }^{-1}}$, finally we will re-substitute the values of a and b which we assumed initially and get our answer.

Complete step-by-step solution:
Let us first take the Left Hand Side of the equation we have: ${{\sin }^{-1}}\left( \dfrac{5}{13} \right)+{{\cos }^{-1}}\left( \dfrac{3}{5} \right)$
Now, let $a={{\sin }^{-1}}\left( \dfrac{5}{13} \right)$ and $b={{\cos }^{-1}}\left( \dfrac{3}{5} \right)$ .......... Equation 1.
Let’s deal with them one by one and find out $\tan a\text{ and }\tan b$ ,
First, we will consider: $a={{\sin }^{-1}}\left( \dfrac{5}{13} \right)$ ,
We know that according to the standard inverse property for inverse function that :${{\sin }^{-1}}x=\theta \Rightarrow x=\sin \theta $ ,
Therefore, $a={{\sin }^{-1}}\left( \dfrac{5}{13} \right)\Rightarrow \sin a=\left( \dfrac{5}{13} \right)\text{ }........\text{Equation 2}\text{.}$
Now we know that: ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\Rightarrow \cos \theta =\sqrt{1-{{\sin }^{2}}\theta }$
Therefore: $\cos a=\sqrt{1-{{\sin }^{2}}a}$ , we will now put the value of $\sin a$ from equation 2:
 $\begin{align}
  & \cos a=\sqrt{1-{{\sin }^{2}}a}\Rightarrow \cos a=\sqrt{1-{{\left( \dfrac{5}{13} \right)}^{2}}} \\
 & \cos a=\sqrt{\dfrac{169-25}{169}}=\sqrt{\dfrac{144}{169}} \\
 & \cos a=\dfrac{12}{13} \\
\end{align}$
Now we know that: $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ ,

Therefore,
 $\tan a=\dfrac{\sin a}{\cos a}$ , we will now be putting the values of $\sin a\text{ and }\cos a$ into this:
$\tan a=\dfrac{\left( \dfrac{5}{13} \right)}{\left( \dfrac{12}{13} \right)}=\left( \dfrac{5}{12} \right)\Rightarrow \tan a=\left( \dfrac{5}{12} \right)\text{ }.........\text{Equation 3}\text{.}$
Similarly we will find $\tan b$ :
Now, we will consider: $b={{\cos }^{-1}}\left( \dfrac{3}{5} \right)$ ,
We know that according to the standard inverse property for inverse function that :${{\cos }^{-1}}x=\theta \Rightarrow x=\cos \theta $ ,
Therefore, $b={{\cos }^{-1}}\left( \dfrac{3}{5} \right)\Rightarrow \cos b=\left( \dfrac{3}{5} \right)\text{ }........\text{Equation 4}\text{.}$
Now we know that: ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\Rightarrow \sin \theta =\sqrt{1-{{\cos }^{2}}\theta }$
Therefore: $\sin b=\sqrt{1-{{\cos }^{2}}b}$ , we will now put the value of $\cos b$ from equation 4:
 $\begin{align}
  & \sin b=\sqrt{1-{{\cos }^{2}}b}\Rightarrow \sin b=\sqrt{1-{{\left( \dfrac{3}{5} \right)}^{2}}} \\
 & \sin b=\sqrt{\dfrac{25-9}{25}}=\sqrt{\dfrac{16}{25}} \\
 & \sin b=\dfrac{4}{5} \\
\end{align}$
Now we know that: $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ ,

Therefore,
 $\tan b=\dfrac{\sin b}{\cos b}$, we will now be putting the values of $\sin b\text{ and }\cos b$ into this:
$\tan b=\dfrac{\left( \dfrac{4}{5} \right)}{\left( \dfrac{3}{5} \right)}=\left( \dfrac{4}{3} \right)\Rightarrow \tan b=\left( \dfrac{4}{3} \right)\text{ }.........\text{Equation 5}\text{.}$
Now we have with us: $\tan a=\dfrac{5}{12}\text{ and }\tan b=\dfrac{4}{3}$ ,
We know the trigonometric property that : $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ , We will apply this property to the values obtained that is $\tan a=\dfrac{5}{12}\text{ and }\tan b=\dfrac{4}{3}$ ,
Now ,
\[\begin{align}
  & \tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}=\dfrac{\dfrac{5}{12}+\dfrac{4}{3}}{1-\left( \dfrac{5}{12}\times \dfrac{4}{3} \right)} \\
 & \tan \left( a+b \right)=\dfrac{\dfrac{\left( 5\times 3 \right)+\left( 4\times 12 \right)}{12\times 3}}{1-\left( \dfrac{20}{36} \right)}=\dfrac{\dfrac{15+48}{36}}{\dfrac{36-20}{36}}=\dfrac{\dfrac{63}{36}}{\dfrac{16}{36}} \\
 & \tan \left( a+b \right)=\dfrac{63}{16} \\
\end{align}\]
Therefore, \[\tan \left( a+b \right)=\dfrac{63}{16}\], we already know that according to the standard inverse property for inverse function that :${{\tan }^{-1}}x=\theta \Rightarrow x=\tan \theta $
So, \[\tan \left( a+b \right)=\dfrac{63}{16}\Rightarrow \left( a+b \right)={{\tan }^{-1}}\left( \dfrac{63}{16} \right)\],
Putting the values of $a\text{ and }b$ from equation 1 that is $a={{\sin }^{-1}}\left( \dfrac{5}{13} \right)$ and $b={{\cos }^{-1}}\left( \dfrac{3}{5} \right)$ ,
in \[\left( a+b \right)={{\tan }^{-1}}\left( \dfrac{63}{16} \right)\],
We will get: ${{\sin }^{-1}}\left( \dfrac{5}{13} \right)+{{\cos }^{-1}}\left( \dfrac{3}{5} \right)={{\tan }^{-1}}\left( \dfrac{63}{16} \right)$ .
Since, L.H.S.=R.H.S
Hence Proved.

Note: Whenever the question comes in the form of sin, cos, and tan, always try and convert the other trigonometric expression in tan form as $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ so the calculation becomes easy this way. Always tell the property that you are using before applying it, do not directly jump into applying it.