Question

Prove that: ${{\cos }^{-1}}\dfrac{63}{65}+2{{\tan }^{-1}}\dfrac{1}{5}={{\sin }^{-1}}\dfrac{3}{5}$ .

Answer 1

Hint: For answering this question we will simplify the left hand side of the given expression and compare it with right hand side and show that they both are equal using the four formulae ${{\cos }^{-1}}\dfrac{a}{b}={{\tan }^{-1}}\dfrac{\sqrt{{{b}^{2}}-{{a}^{2}}}}{a}$ and $2{{\tan }^{-1}}a={{\tan }^{-1}}\dfrac{2a}{1-{{a}^{2}}}$ and ${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\dfrac{a+b}{1-ab}$ and ${{\tan }^{-1}}\dfrac{a}{b}={{\sin }^{-1}}\dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$ .

Complete step by step answer:
Now considering from the question we have to prove ${{\cos }^{-1}}\dfrac{63}{65}+2{{\tan }^{-1}}\dfrac{1}{5}={{\sin }^{-1}}\dfrac{3}{5}$ .
From the basic concept we have ${{\cos }^{-1}}\dfrac{a}{b}={{\tan }^{-1}}\dfrac{\sqrt{{{b}^{2}}-{{a}^{2}}}}{a}$ we will use this and simplify the left hand side expression then we will have ${{\cos }^{-1}}\dfrac{63}{65}={{\tan }^{-1}}\dfrac{\sqrt{{{65}^{2}}-{{63}^{2}}}}{63}$ .
By simplifying this we will have
${{\cos }^{-1}}\dfrac{63}{65}={{\tan }^{-1}}\dfrac{\sqrt{4225-3969}}{63}\Rightarrow {{\cos }^{-1}}\dfrac{63}{65}={{\tan }^{-1}}\dfrac{\sqrt{256}}{63}\Rightarrow {{\cos }^{-1}}\dfrac{63}{65}={{\tan }^{-1}}\dfrac{16}{63}$ .
After simplifying we will have ${{\tan }^{-1}}\dfrac{16}{63}+2{{\tan }^{-1}}\dfrac{1}{5}$ .
From the basic concept we have $2{{\tan }^{-1}}a={{\tan }^{-1}}\dfrac{2a}{1-{{a}^{2}}}$ we will use this and simplify the expression further we will have $2{{\tan }^{-1}}\dfrac{1}{5}={{\tan }^{-1}}\dfrac{2\left( \dfrac{1}{5} \right)}{1-\dfrac{1}{25}}\Rightarrow 2{{\tan }^{-1}}\dfrac{1}{5}={{\tan }^{-1}}\dfrac{5}{12}$ .
After this we will have ${{\tan }^{-1}}\dfrac{16}{63}+{{\tan }^{-1}}\dfrac{5}{12}$ .
As we know that from the basic concept that we have a formulae stating ${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\dfrac{a+b}{1-ab}$ after using this in the expression we will have${{\tan }^{-1}}\dfrac{16}{63}+{{\tan }^{-1}}\dfrac{5}{12}={{\tan }^{-1}}\dfrac{\dfrac{16}{63}+\dfrac{5}{12}}{1-\dfrac{16}{63}\dfrac{5}{12}}\Rightarrow {{\tan }^{-1}}\dfrac{16\times 12+63\times 5}{63\times 12-5\times 16}={{\tan }^{-1}}\dfrac{507}{676}={{\tan }^{-1}}\dfrac{3}{4}$ .
By using this in the expression we will have ${{\tan }^{-1}}\dfrac{3}{4}$ .
Now we will use the formulae ${{\tan }^{-1}}\dfrac{a}{b}={{\sin }^{-1}}\dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$ and simplify it after that we will have${{\tan }^{-1}}\dfrac{3}{4}={{\sin }^{-1}}\dfrac{3}{\sqrt{{{3}^{2}}+{{4}^{2}}}}={{\sin }^{-1}}\dfrac{3}{5}$ .
Hence we can observe that the left hand side of the expression is equal to its right hand side.
So it has been proved that the expression ${{\cos }^{-1}}\dfrac{63}{65}+2{{\tan }^{-1}}\dfrac{1}{5}={{\sin }^{-1}}\dfrac{3}{5}$ is valid.

Note: While answering questions of this type we should take care of the calculations like in the case of this question if we had made a mistake and taken ${{\tan }^{-1}}\dfrac{507}{676}={{\tan }^{-1}}\dfrac{3}{5}$ we will get an answer as ${{\tan }^{-1}}\dfrac{3}{5}={{\sin }^{-1}}\dfrac{3}{\sqrt{{{3}^{2}}+{{5}^{2}}}}={{\sin }^{-1}}\dfrac{3}{\sqrt{34}}$ so we will be unable to prove the given expression.