Question

  In any \[\Delta ABC\], prove that
\[\left( {{c}^{2}}-{{a}^{2}}+{{b}^{2}} \right)\tan A=\left( {{a}^{2}}-{{b}^{2}}+{{c}^{2}} \right)\tan B=\left( {{b}^{2}}-{{c}^{2}}+{{a}^{2}} \right)\tan C\]

Answer 1

Hint: For the above question, we will use the sine rule and cosine rule of the triangle and we know that the tangent of any angle is the ratio of the sine to the cosine of that angle. So, we will substitute the value of sine and cosine from the rule to the expression and we will get all the expressions in a constant term.

Complete step by step answer:
We have been given to prove
\[\left( {{c}^{2}}-{{a}^{2}}+{{b}^{2}} \right)\tan A=\left( {{a}^{2}}-{{b}^{2}}+{{c}^{2}} \right)\tan B=\left( {{b}^{2}}-{{c}^{2}}+{{a}^{2}} \right)\tan C\]
\[\left( {{c}^{2}}-{{a}^{2}}+{{b}^{2}} \right)\dfrac{\sin A}{\cos A}=\left( {{a}^{2}}-{{b}^{2}}+{{c}^{2}} \right)\dfrac{\sin B}{\cos B}=\left( {{b}^{2}}-{{c}^{2}}+{{a}^{2}} \right)\dfrac{\sin C}{\cos C}\]
We know that the sine rule is as follows:
\[\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}\]
Let the ratio is equal to ‘k’, we get,
\[\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=k\]
\[\Rightarrow \sin A=ak,\text{ }\sin B=bk\text{ and }\sin C=ck\]
Let us take the expression as follows:
\[\left( {{c}^{2}}-{{a}^{2}}+{{b}^{2}} \right)\dfrac{\sin A}{\cos A}....\left( i \right)\]
Also, we have a cosine rule as follows:
\[\cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\]
\[\cos B=\dfrac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\]
\[\cos C=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}\]
Substituting the values of sin A and cos A in the expression (i), we get,
\[\Rightarrow \left( {{c}^{2}}-{{a}^{2}}+{{b}^{2}} \right)\dfrac{\sin A}{\cos A}=\left( {{c}^{2}}-{{a}^{2}}+{{b}^{2}} \right)\dfrac{ka\times 2bc}{\left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}} \right)}\]
\[\Rightarrow \left( {{c}^{2}}-{{a}^{2}}+{{b}^{2}} \right)\dfrac{\sin A}{\cos A}=2\left( abc \right)k\]
Similarly, substituting the values of sin B and cos B in the other expression, we get,
\[\Rightarrow \left( {{a}^{2}}-{{b}^{2}}+{{c}^{2}} \right)\dfrac{\sin B}{\cos B}=\left( {{a}^{2}}-{{b}^{2}}+{{c}^{2}} \right)\dfrac{kb\times 2ac}{\left( {{a}^{2}}+{{c}^{2}}-{{b}^{2}} \right)}\]
\[\Rightarrow \left( {{a}^{2}}-{{b}^{2}}+{{c}^{2}} \right)\dfrac{\sin B}{\cos B}=2\left( abc \right)k\]
Again, substituting the values of sin C and cos C in the other expression, we get,
\[\Rightarrow \left( {{b}^{2}}-{{c}^{2}}+{{a}^{2}} \right)\dfrac{\sin C}{\cos C}=\left( {{b}^{2}}-{{c}^{2}}+{{a}^{2}} \right)\dfrac{kc\times 2ab}{\left( {{a}^{2}}+{{b}^{2}}-{{c}^{2}} \right)}\]
\[\Rightarrow \left( {{b}^{2}}-{{c}^{2}}+{{a}^{2}} \right)\dfrac{\sin C}{\cos C}=2\left( abc \right)k\]
Since we get all the expression as equal to 2(abc)k, hence we have proved that
\[\left( {{c}^{2}}-{{a}^{2}}+{{b}^{2}} \right)\tan A=\left( {{a}^{2}}-{{b}^{2}}+{{c}^{2}} \right)\tan B=\left( {{b}^{2}}-{{c}^{2}}+{{a}^{2}} \right)\tan C\]

Note: We can easily solve the above question if we know the formula of \[\tan A=\dfrac{abc}{R\left( {{b}^{2}}+{{c}^{2}}-{{a}^{2}} \right)}\] where R is the circumradius of \[\Delta ABC\] and remember the Basic formulas of trigonometric.