Question

Prove the following result ${{\tan }^{-1}}\left( \dfrac{2ab}{{{a}^{2}}-{{b}^{2}}} \right)+{{\tan }^{-1}}\left( \dfrac{2xy}{{{x}^{2}}-{{y}^{2}}} \right)={{\tan }^{-1}}\left( \dfrac{2\alpha \beta }{{{\alpha }^{2}}-{{\beta }^{2}}} \right)$.

Answer 1

Hint: In order to find the solution of this question, we should have some knowledge about the inverse trigonometric formulas like ${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$. Also, we should know a few algebraic identities like, ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ and ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. By using these formulas, we can prove the desired result.

Complete step-by-step answer:
In this question, we have been asked to prove that ${{\tan }^{-1}}\left( \dfrac{2ab}{{{a}^{2}}-{{b}^{2}}} \right)+{{\tan }^{-1}}\left( \dfrac{2xy}{{{x}^{2}}-{{y}^{2}}} \right)={{\tan }^{-1}}\left( \dfrac{2\alpha \beta }{{{\alpha }^{2}}-{{\beta }^{2}}} \right)$. So, to prove this, we will first consider the left hand side of the given equality. So, we can write it as,
$LHS={{\tan }^{-1}}\left( \dfrac{2ab}{{{a}^{2}}-{{b}^{2}}} \right)+{{\tan }^{-1}}\left( \dfrac{2xy}{{{x}^{2}}-{{y}^{2}}} \right)$
Now, we know that ${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right)$. So, for $a=\dfrac{2ab}{{{a}^{2}}-{{b}^{2}}}$ and $b=\dfrac{2xy}{{{x}^{2}}-{{y}^{2}}}$, we can write the LHS as,
\[LHS={{\tan }^{-1}}\left[ \dfrac{\dfrac{2ab}{{{a}^{2}}-{{b}^{2}}}+\dfrac{2xy}{{{x}^{2}}-{{y}^{2}}}}{1-\left( \dfrac{2ab}{{{a}^{2}}-{{b}^{2}}} \right)\left( \dfrac{2xy}{{{x}^{2}}-{{y}^{2}}} \right)} \right]\]
Now, we will take LCM of both the terms in the numerator and in the denominator. So, we will get,
\[LHS={{\tan }^{-1}}\left[ \dfrac{\dfrac{2ab\left( {{x}^{2}}-{{y}^{2}} \right)+2xy\left( {{a}^{2}}-{{b}^{2}} \right)}{\left( {{a}^{2}}-{{b}^{2}} \right)\left( {{x}^{2}}-{{y}^{2}} \right)}}{\dfrac{\left( {{a}^{2}}-{{b}^{2}} \right)\left( {{x}^{2}}-{{y}^{2}} \right)-\left( 2ab \right)\left( 2xy \right)}{\left( {{a}^{2}}-{{b}^{2}} \right)\left( {{x}^{2}}-{{y}^{2}} \right)}} \right]\]
We can further write it as,
\[LHS={{\tan }^{-1}}\left[ \dfrac{\left[ 2ab\left( {{x}^{2}}-{{y}^{2}} \right)+2xy\left( {{a}^{2}}-{{b}^{2}} \right) \right]\left[ \left( {{a}^{2}}-{{b}^{2}} \right)\left( {{x}^{2}}-{{y}^{2}} \right) \right]}{\left[ \left( {{a}^{2}}-{{b}^{2}} \right)\left( {{x}^{2}}-{{y}^{2}} \right)-\left( 2ab \right)\left( 2xy \right) \right]\left[ \left( {{a}^{2}}-{{b}^{2}} \right)\left( {{x}^{2}}-{{y}^{2}} \right) \right]} \right]\]
Now, we know that the common terms of the numerator and the denominator will get cancelled out. So, we can write the LHS as,
\[LHS={{\tan }^{-1}}\left[ \dfrac{2ab\left( {{x}^{2}}-{{y}^{2}} \right)+2xy\left( {{a}^{2}}-{{b}^{2}} \right)}{\left( {{a}^{2}}-{{b}^{2}} \right)\left( {{x}^{2}}-{{y}^{2}} \right)-\left( 2ab \right)\left( 2xy \right)} \right]\]
Now, we will open the brackets in order to simplify it, so we will get,
\[LHS={{\tan }^{-1}}\left[ \dfrac{2ab{{x}^{2}}-2ab{{y}^{2}}+2xy{{a}^{2}}-2xy{{b}^{2}}}{{{a}^{2}}{{x}^{2}}-{{b}^{2}}{{x}^{2}}-{{a}^{2}}{{y}^{2}}+{{b}^{2}}{{y}^{2}}-4abxy} \right]\]
Now, we know that 4 abxy = 2 abxy + 2 abxy. So, we will get the LHS as,
\[LHS={{\tan }^{-1}}\left[ \dfrac{2ab{{x}^{2}}-2ab{{y}^{2}}+2xy{{a}^{2}}-2xy{{b}^{2}}}{{{a}^{2}}{{x}^{2}}-{{b}^{2}}{{x}^{2}}-{{a}^{2}}{{y}^{2}}+{{b}^{2}}{{y}^{2}}-2abxy-2abxy} \right]\]
In numerator, we can see that 2 (ax) can be taken out as the common term from $\left( 2ab{{x}^{2}}+2xy{{a}^{2}} \right)$ and – (2by) can be taken out as the common term from $\left( -2ab{{y}^{2}}-2xy{{b}^{2}} \right)$. Therefore, we will get,
\[LHS={{\tan }^{-1}}\left[ \dfrac{2ax\left( bx+ay \right)-2by\left( bx+ay \right)}{\left( {{a}^{2}}{{x}^{2}}-{{b}^{2}}{{y}^{2}}-2abxy \right)-\left( {{a}^{2}}{{y}^{2}}+{{b}^{2}}{{x}^{2}}+2abxy \right)} \right]\]
Now, we can see that 2 (bx + ay) can be taken out as common from the numerator. Also, we know that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ and ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. So, we can write \[\left( {{a}^{2}}{{x}^{2}}-{{b}^{2}}{{y}^{2}}-2abxy \right)\] as ${{\left( ax-by \right)}^{2}}$ and we can write \[\left( {{a}^{2}}{{y}^{2}}+{{b}^{2}}{{x}^{2}}+2abxy \right)\] as ${{\left( ay+bx \right)}^{2}}$. So, we get the LHS as,
\[LHS={{\tan }^{-1}}\left[ \dfrac{2\left( bx+ay \right)\left( ax-by \right)}{{{\left( ax-by \right)}^{2}}-{{\left( bx+ay \right)}^{2}}} \right]\]
If we consider $\left( ax-by \right)=\alpha $ and $\left( bx+ay \right)=\beta $, then we can say that,
\[LHS={{\tan }^{-1}}\left( \dfrac{2\alpha \beta }{{{\alpha }^{2}}-{{\beta }^{2}}} \right)\]
LHS = RHS
Hence proved.

Note: While solving this question, one can think that if we have the LHS in terms of a, b, x and y, then how can we get the RHS in terms of α and β. For that we have to find α and β in terms of a, b, x and y, then we have to represent them in terms of \[{{\tan }^{-1}}\left( \dfrac{2\alpha \beta }{{{\alpha }^{2}}-{{\beta }^{2}}} \right)\].