Question

If ${{\tan }^{-1}}\dfrac{x-1}{x-2}+{{\tan }^{-1}}\dfrac{x+1}{x+2}=\dfrac{\pi }{4}$ then find the value of $x.$

Answer 1

Hint: As we have the given expression there is ${{\tan }^{-1}}$ and we have to determine the value of $x.$ We know that the ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left[ \dfrac{x+y}{1-xy} \right]$, Now we have to apply this in the given the expression for that we have to replace the $\dfrac{x-1}{x-2}$ by $x$ and $\dfrac{x+1}{x+2}$ by $y.$ Then you have to solve right hand side with the help of cross multiplication solved in numerator and denominator then we have to get the bracket terms solved. But at some point we have to use the identity of $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ After that we can easily find the value of $x.$

Complete step-by-step solution:
The given expression is,
${{\tan }^{-1}}\dfrac{x-1}{x-2}+{{\tan }^{-1}}\dfrac{x+1}{x+2}=\dfrac{\pi }{4}$
Now we know that, the formula of
${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left[ \dfrac{x+y}{1-xy} \right]$
After that we have to replace the $x$ by $\dfrac{x+1}{x+2}$ in given from the expression.
Substitute given values in the formula get,
${{\tan }^{-1}}\left( \dfrac{x-1}{x+2} \right)+{{\tan }^{-1}}\left( \dfrac{x+1}{x+2} \right)$
$={{\tan }^{-1}}\left[ \dfrac{\dfrac{x-1}{x+2}+\dfrac{x+1}{x+2}}{1-\dfrac{x-1}{x+2}\times \dfrac{x+1}{x+2}} \right]$
$\therefore {{\tan }^{-1}}\left[ \dfrac{x+1}{x+2} \right]+{{\tan }^{-1}}\left[ \dfrac{x+1}{x+2} \right]={{\tan }^{-1}}\left[ \dfrac{\dfrac{\left( x+1 \right)\left( x+2 \right)+\left( x+1 \right)\left( x+2 \right)}{\left( x-1 \right)\left( x+2 \right)}}{\dfrac{\left( x-2 \right)\left( x+2 \right)-\left( x-1 \right)\left( x+1 \right)}{\left( x-2 \right)\left( x+2 \right)}} \right]$ [By cross multiplication]
$={{\tan }^{-1}}\left[ \dfrac{\left( x-1 \right)\left( x+2 \right)+\left( x+1 \right)\left( x-2 \right)}{\left( x-2 \right)\left( x+2 \right)}\times \dfrac{\left( x-2 \right)\left( x+2 \right)}{\left( x+2 \right)\left( x-2 \right)-\left( x-1 \right)\left( x+1 \right)} \right]$
Simplifies it.
$={{\tan }^{-1}}\left[ \dfrac{\left( x-1 \right)\left( x+2 \right)+\left( x+1 \right)\left( x-2 \right)}{\left( x+2 \right)\left( x-2 \right)-\left( x-1 \right)\left( x+1 \right)} \right]$
As we know that the identity of $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$

By using the identity we get,
$={{\tan }^{-1}}\left[ \dfrac{\left( x-1 \right)\left( x+2 \right)+\left( x+1 \right)\left( x-2 \right)}{{{x}^{2}}-{{2}^{2}}-\left[ {{x}^{2}}-{{1}^{2}} \right]} \right]$
Solve it,
$={{\tan }^{-1}}\left[ \dfrac{x\left( x+2 \right)-1\left( x+2 \right)+x\left( x+2 \right)+1\left( x-2 \right)}{{{x}^{2}}-4-{{x}^{2}}+1} \right]$
$={{\tan }^{-1}}\left[ \dfrac{{{x}^{2}}+2x-2+{{x}^{2}}-2x+x-2}{{{x}^{2}}-4-{{x}^{2}}+1} \right]$
Cancel out the similar terms we get,
$={{\tan }^{-1}}\left[ \dfrac{2{{x}^{2}}-4}{-3} \right]$
Therefore, given expression is written as,
${{\tan }^{-1}}\left[ \dfrac{x-1}{x-2} \right]+{{\tan }^{-1}}\left[ \dfrac{x+1}{x+2} \right]={{\tan }^{-1}}\left[ \dfrac{2{{x}^{2}}-4}{-3} \right]...(i)$
Here given,
${{\tan }^{-1}}\left[ \dfrac{x-1}{x-2} \right]+{{\tan }^{-1}}\left[ \dfrac{x+1}{x+2} \right]=\dfrac{\pi }{4}$
Substitute value of equation $(i)$
$\therefore {{\tan }^{-1}}\left[ \dfrac{2{{x}^{2}}-4}{-3} \right]=\dfrac{\pi }{4}$
Transfer the ${{\tan }^{-1}}$ to the right side.
$\left[ \dfrac{2{{x}^{2}}-4}{-3} \right]=\tan \dfrac{\pi }{4}$
$\left[ \dfrac{2{{x}^{2}}-4}{-3} \right]=1$ Because $\left[ \because \tan \dfrac{\pi }{4}=1 \right]$
Solve it,
$\therefore 2{{x}^{2}}-4=-3$
$\therefore 2{{x}^{2}}=-3+4$
$\therefore 2{{x}^{2}}=1$
${{x}^{2}}=\dfrac{1}{2}$
$x=\pm \dfrac{1}{\sqrt{2}}$

The value of $x$ is $\pm \dfrac{1}{\sqrt{2}}$

Note: We have to use the sum and difference identities for sine, cosine and tangent while solving any trigonometric problems you need to convert all sec, cot csc and to sin and cos. This helps us to solve it easily by using identities. This checks all sum and differences.