Given:
[tex]\tan A=\dfrac{1}{5}[/tex]
[tex]\tan B=\dfrac{1}{4}[/tex]
To find:
The value of [tex]\tan^2(A-B)[/tex].
Solution:
We know that,
[tex]\tan (A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}[/tex]
Putting the given values, we get
[tex]\tan (A-B)=\dfrac{\dfrac{1}{5}-\dfrac{1}{4}}{1+\dfrac{1}{5}\cdot \dfrac{1}{4}}[/tex]
[tex]\tan (A-B)=\dfrac{\dfrac{4-5}{20}}{1+\dfrac{1}{20}}[/tex]
[tex]\tan (A-B)=\dfrac{\dfrac{-1}{20}}{\dfrac{20+1}{20}}[/tex]
[tex]\tan (A-B)=\dfrac{-1}{21}[/tex]
Taking square on both sides, we get
[tex]\tan^2 (A-B)=\left( \dfrac{-1}{21}\right)^2[/tex]
[tex]\tan^2 (A-B)=\dfrac{1}{441}[/tex]
Therefore, the value of [tex]\tan^2(A-B)[/tex] is [tex]\dfrac{1}{441}[/tex].
