VerifiedHint: In the question, LHS has inverse sine functions and RHS has inverse cosine function. We have inverse cosine functions in the RHS of the given expression and inverse sine function in the LHS of the given expression. So first we will try to transform them. We can do so by taking the term \[x={{\sin }^{-1}}\dfrac{3}{5}\] and \[y={{\sin }^{-1}}\dfrac{8}{17}\]. Then, we can apply the formula \[\cos (x-y)=cosx.cosy+sinx.siny\] to simplify and solve further.
Complete step by step answer:
Now, we have to remove the inverse functions from the terms present in LHS.
Let us assume,
\[x=si{{n}^{-1}}\dfrac{3}{5}\] ………….. (1)
The RHS part in the question is given in inverse cosine function. So, we need to convert this inverse cosine into a cosine function.
Now, applying sine in LHS as well as RHS in the equation,
\[ x=si{{n}^{-1}}\dfrac{3}{5} \]
\[ \Rightarrow \sin x=\dfrac{3}{5} \]
Also, we have to convert the sine function into cosine.
We know the identity \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\] ,using this identity we get
\[\cos x=\sqrt{1-{{\sin }^{2}}x}\] ………….(2)
Putting the value of “sin x” in the equation (2), we get
\[\cos x=\sqrt{1-{{\sin }^{2}}x}\]
\[ \cos x=\sqrt{1-{{\left( \dfrac{3}{5} \right)}^{2}}} \]
\[ \Rightarrow \cos x=\sqrt{1-\dfrac{9}{25}} \]
\[ \Rightarrow \cos x=\sqrt{\dfrac{25-9}{25}} \]
\[ \Rightarrow \cos x=\sqrt{\dfrac{16}{25}} \]
\[ \Rightarrow \cos x=\dfrac{4}{5} \]
Similarly, let us assume,
\[y=si{{n}^{-1}}\dfrac{8}{17}\]………….. (3)
Now, applying sine in LHS as well as RHS in the equation,
\[ y=si{{n}^{-1}}\dfrac{8}{17} \]
\[ \Rightarrow \sin y=\dfrac{8}{17} \]
Also, we have to convert the sine function into cosine.
We know that, \[\operatorname{cosy}=\sqrt{1-{{\sin }^{2}}y}\] .
Putting the value of “sin y” in the equation, , we get
\[\operatorname{cosy}=\sqrt{1-{{\sin }^{2}}y}\]
\[ \operatorname{cosy}=\sqrt{1-{{\left( \dfrac{8}{17} \right)}^{2}}} \]
\[ \Rightarrow \operatorname{cosy}=\sqrt{1-\dfrac{64}{289}} \]
\[ \Rightarrow \operatorname{cosy}=\sqrt{\dfrac{289-64}{289}} \]
\[ \Rightarrow \operatorname{cosy}=\sqrt{\dfrac{225}{289}} \]
\[ \Rightarrow \operatorname{cosy}=\dfrac{15}{17} \]
Putting the equation(1) and equation (3) in the given expression, we get,
\[ {{\sin }^{-1}}\dfrac{3}{5}-{{\sin }^{-1}}\dfrac{8}{17} \]
\[ =x-y \]
This means we have to find (x-y).
As in RHS, we have cosine terms, so using the formula expand cos(x-y).
\[ \cos (x-y)=cosx.cosy-sinx.siny \]
\[ \cos (x-y)=\dfrac{4}{5}.\dfrac{15}{17}+\dfrac{3}{5}.\dfrac{8}{17} \]
\[ \cos (x-y)=\dfrac{84}{85} \]
\[ (x-y)=co{{s}^{-1}}\dfrac{84}{85}. \]
Therefore, LHS=RHS.
Hence, proved.
Note: We can also solve this question, after the conversion of the inverse of cosine present in LHS as well as RHS into the inverse of tan. We can do so by taking \[{{\sin }^{-1}}(3/5)=x\] and \[{{\sin }^{-1}}(8/17)=y\].
\[ {{\sin }^{-1}}\left( \dfrac{3}{5} \right)=x \]
\[ \Rightarrow sinx=\dfrac{3}{5} \]
Using the Pythagoras theorem, we can find the perpendicular.
\[Perpendicular=\sqrt{{{\left( hypotenuse \right)}^{2}}-{{\left( base \right)}^{2}}}\]
\[ =\sqrt{{{5}^{2}}-{{3}^{2}}} \]
\[ =\sqrt{25-9} \]
\[ =\sqrt{16} \]
\[ =4 \]
\[\tan x=\dfrac{Perpendicular}{Base}\]
\[\tan x=\dfrac{3}{4}\]
Similarly,
\[ si{{n}^{-1}}\left( \dfrac{8}{17} \right)=y \]
\[ \Rightarrow siny=\dfrac{8}{17} \]
Using the Pythagoras theorem, we can find the base.
\[Base=\sqrt{{{\left( Hypotenuse \right)}^{2}}-{{\left( Perpendicular \right)}^{2}}}\]
\[ =\sqrt{{{17}^{2}}-{{8}^{2}}} \]
\[ =\sqrt{289-64} \]
\[ =\sqrt{225} \]
\[ =15 \]
\[\tan x=\dfrac{Perpendicular}{Base}\]
\[\tan x=\dfrac{8}{15}\]
And then using the formula \[{{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)\] , we can get the required result.
