Question

Prove the following: ${{\cos }^{-1}}\left( \dfrac{12}{13} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)={{\sin }^{-1}}\left( \dfrac{56}{65} \right)$

Answer 1

Hint: In order to solve these types of question, we have to convert both the terms of LHS in ${{\sin }^{-1}}\theta $ and then compare the value with RHS, if both are equal then statement is proved. In order to convert ${{\cos }^{-1}}\left( \dfrac{12}{13} \right)$ in terms of ${{\sin }^{-1}}\theta $ use the formula $\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }$ and then apply the sum rule of ${{\sin }^{-1}}\theta $ which is ${{\sin }^{-1}}a+{{\sin }^{-1}}b={{\sin }^{-1}}\left\{ a\sqrt{1-{{b}^{2}}}+b\sqrt{1-{{a}^{2}}} \right\}$ and hence verify the statement.

Complete step by step answer:
It is given that LHS is equal to ${{\cos }^{-1}}\left( \dfrac{12}{13} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)$.
Now, we have to convert ${{\cos }^{-1}}\left( \dfrac{12}{13} \right)$ in terms of ${{\sin }^{-1}}\theta $.
Let us suppose that $x=~{{\cos }^{-1}}\left( \dfrac{12}{13} \right)$
Then, $\cos x=\dfrac{12}{13}$
As we know that $\sin \theta =\sqrt{1-{{\cos }^{2}}\theta }$. So, using this formula, we get
$\therefore \sin x=\sqrt{1-{{\dfrac{12}{13}}^{2}}}=\sqrt{1-\dfrac{144}{169}}=\sqrt{\dfrac{25}{169}}=\dfrac{5}{13}$
We can also write the above expression as $x={{\sin }^{-1}}\left( \dfrac{5}{13} \right)$
Now, $x={{\sin }^{-1}}\left( \dfrac{5}{13} \right)={{\cos }^{-1}}\left( \dfrac{12}{13} \right)\Rightarrow {{\sin }^{-1}}\left( \dfrac{5}{13} \right)={{\cos }^{-1}}\left( \dfrac{12}{13} \right)..................(1)$

We have,
$\therefore {{\cos }^{-1}}\left( \dfrac{12}{13} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)$
We know that from equation (1), ${{\sin }^{-1}}\left( \dfrac{5}{13} \right)={{\cos }^{-1}}\left( \dfrac{12}{13} \right)$ . So, substituting ${{\sin }^{-1}}\left( \dfrac{5}{13} \right)$ in place of ${{\cos }^{-1}}\left( \dfrac{12}{13} \right)$ in the above expression, we get
\[\Rightarrow {{\sin }^{-1}}\left( \dfrac{5}{13} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)\]
We also know that ${{\sin }^{-1}}a+{{\sin }^{-1}}b={{\sin }^{-1}}\left\{ a\sqrt{1-{{b}^{2}}}+b\sqrt{1-{{a}^{2}}} \right\}$, using this relation in the above expression and substituting values in place of $a$ and $b$, we get
In our case $a=\dfrac{5}{13}$ and $b=\dfrac{3}{5}$.
$\Rightarrow {{\sin }^{-1}}\left\{ \dfrac{5}{13}\sqrt{1-{{\left( \dfrac{3}{5} \right)}^{2}}}+\dfrac{3}{5}\sqrt{1-{{\left( \dfrac{5}{13} \right)}^{2}}} \right\}$
$\Rightarrow {{\sin }^{-1}}\left\{ \dfrac{5}{13}\sqrt{1-\dfrac{9}{25}}+\dfrac{3}{5}\sqrt{1-\dfrac{25}{169}} \right\}$
$\Rightarrow {{\sin }^{-1}}\left\{ \dfrac{5}{13}\sqrt{\dfrac{16}{25}}+\dfrac{3}{5}\sqrt{\dfrac{144}{169}} \right\}$
$\Rightarrow {{\sin }^{-1}}\left\{ \dfrac{5}{13}\times \dfrac{4}{5}+\dfrac{3}{5}\times \dfrac{12}{13} \right\}$
$\Rightarrow {{\sin }^{-1}}\left\{ \dfrac{20}{65}+\dfrac{36}{65} \right\}={{\sin }^{-1}}\left( \dfrac{56}{65} \right)$
Hence, the value of expression ${{\cos }^{-1}}\left( \dfrac{12}{13} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)$ is equal to ${{\sin }^{-1}}\left( \dfrac{56}{65} \right)$.
So, ${{\cos }^{-1}}\left( \dfrac{12}{13} \right)+{{\sin }^{-1}}\left( \dfrac{3}{5} \right)$ = ${{\sin }^{-1}}\left( \dfrac{56}{65} \right)$ = RHS
Hence, as LHS=RHS so above given statement is proved.

Note: This type of question involves calculating and substituting values in the expression with extreme precision. Students should carefully convert one expression into the other like the way we did in the above expression where ${{\sin }^{-1}}\left( \dfrac{5}{13} \right)={{\cos }^{-1}}\left( \dfrac{12}{13} \right)$. Students should try to match the expression in RHS and LHS as it reduces the total calculation like in our case we have ${{\sin }^{-1}}\theta $ and ${{\cos }^{-1}}\theta $ in the LHS and in RHS we have ${{\sin }^{-1}}\theta $. So, we converted both terms of LHS in ${{\sin }^{-1}}\theta $ and proved the statement.