Question

Prove that \[{{\cos }^{-1}}\left( \dfrac{12}{13} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)={{\sin }^{-1}}\left( \dfrac{56}{65} \right)\]

Answer 1

Hint: In this question, we first need to consider the left hand side of the given expression and then using the properties of inverse trigonometric functions we need to convert the inverse cosine function into inverse sine function. Then use the sum of the inverse functions formula accordingly and simplify further to get the result.

Complete step-by-step answer:
\[{{\cos }^{-1}}x={{\sin }^{-1}}\sqrt{1-{{x}^{2}}}\]
\[{{\sin }^{-1}}x+{{\sin }^{-1}}y={{\sin }^{-1}}\left\{ x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}} \right\}\]
Now, from the given expression in the question we have
\[\Rightarrow {{\cos }^{-1}}\left( \dfrac{12}{13} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)={{\sin }^{-1}}\left( \dfrac{56}{65} \right)\]
Let us now consider the left hand side of the given expression
\[\Rightarrow {{\cos }^{-1}}\left( \dfrac{12}{13} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)\]
As we already know that from the properties of inverse trigonometric functions the relation between inverse cosine and inverse sine functions can be expressed as follows
\[{{\cos }^{-1}}x={{\sin }^{-1}}\sqrt{1-{{x}^{2}}}\]
Now, from this property on substituting the respective values we can write the expression as
\[\Rightarrow {{\sin }^{-1}}\left( \sqrt{1-{{\left( \dfrac{12}{13} \right)}^{2}}} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)\]
Now, on further simplification we can write it as
\[\Rightarrow {{\sin }^{-1}}\left( \sqrt{\dfrac{169-144}{{{13}^{2}}}} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)\]
Let us now further simplify the above expression
\[\Rightarrow {{\sin }^{-1}}\left( \sqrt{\dfrac{25}{{{13}^{2}}}} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)\]
Now, the above expression can be rewritten as
\[\Rightarrow {{\sin }^{-1}}\left( \dfrac{5}{13} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)\]
As we already know that from the properties of inverse trigonometric functions the sum of two inverse sine functions can be written as follows
\[{{\sin }^{-1}}x+{{\sin }^{-1}}y={{\sin }^{-1}}\left\{ x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}} \right\}\]
Now, on substituting the respective values of x and y in the above equation we get,
Now, on comparison we get,
\[x=\dfrac{5}{13},y=\dfrac{3}{5}\]
Now, on further substitution of these values we get,
\[\Rightarrow {{\sin }^{-1}}\left\{ \dfrac{5}{13}\sqrt{1-{{\left( \dfrac{3}{5} \right)}^{2}}}+\dfrac{3}{5}\sqrt{1-{{\left( \dfrac{5}{13} \right)}^{2}}} \right\}\]
Let us now further simplify this accordingly
\[\Rightarrow {{\sin }^{-1}}\left\{ \dfrac{5}{13}\sqrt{\dfrac{25-9}{{{5}^{2}}}}+\dfrac{3}{5}\sqrt{\dfrac{169-25}{{{13}^{2}}}} \right\}\]
Now, on further simplification we get,
\[\Rightarrow {{\sin }^{-1}}\left\{ \dfrac{5}{13}\sqrt{\dfrac{16}{{{5}^{2}}}}+\dfrac{3}{5}\sqrt{\dfrac{144}{{{13}^{2}}}} \right\}\]
Now, this can be further written as
\[\Rightarrow {{\sin }^{-1}}\left\{ \dfrac{5}{13}\times \dfrac{4}{5}+\dfrac{3}{5}\times \dfrac{12}{13} \right\}\]
Now, on further simplification we get,
\[\begin{align}
  & \Rightarrow {{\sin }^{-1}}\left\{ \dfrac{20}{65}+\dfrac{36}{65} \right\} \\
 & \Rightarrow {{\sin }^{-1}}\left( \dfrac{56}{65} \right) \\
\end{align}\]
Hence, proved that \[{{\cos }^{-1}}\left( \dfrac{12}{13} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)={{\sin }^{-1}}\left( \dfrac{56}{65} \right)\]

Note:Instead of considering the left hand side we can also solve it by first rearranging the terms in the given expression by taking the inverse sine function on the left side to the right side and then solve it by using the formula of difference between two inverse sine functions. Then on further simplification we get the result.
It is important to note that while solving the expression on substituting the respective values we should not neglect any of the terms because it changes the corresponding expression and so the final result.