Question

Prove that $\frac{a^2+b^2-c^2}{c^2+a^2-b^2}=\frac{\tan B}{\tan C}$.

Answer 1

To prove:-

$\frac{a^2+b^2-c^2}{c^2+a^2-b^2}=\frac{\tan B}{\tan C}$

Proof:-

To prove above identity, we will use sine and cosine rule-

Sine rule:-

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k\left(\text{say}\right)\text{OR}\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k\left(\text{say}\right)$

Cosine rule:-

$a^2=b^2+c^2-2bc\cos A$

$b^2=c^2+a^2-2ca\cos B$

$c^2=a^2+b^2-2ab\cos C$

Now

$\frac{a^2+b^2-c^2}{c^2+a^2-b^2}=\frac{\tan B}{\tan C}$

Taking L.H.S.-

$\frac{a^2+b^2-c^2}{c^2+a^2-b^2}$

From cosine rule, we have

$=\frac{a^2+b^2-(a^2+b^2-2ab\cos C)}{c^2+a^2-(a^2+c^2-2ac\cos B)}$

$\Rightarrow =\frac{a^2+b^2-a^2-b^2+2ab\cos C}{c^2+a^2-a^2-c^2+2ac\cos B}$

$\Rightarrow =\frac{2ab\cos C}{2ac\cos B}$

$\Rightarrow =\frac{b\cos C}{c\cos B}$

Now from sine rule, we have

$=\frac{k\sin B\cos C}{k\sin C\cos B}[\because c=k\sin C\&b=k\sin B]$

$=\frac{\left(\frac{\sin B}{\cos B}\right)}{\left(\frac{\sin C}{\cos C}\right)}$

$=\frac{\tan B}{\tan C}$

$=$ R.H.S.

Hence proved that $\frac{a^2+b^2-c^2}{c^2+a^2-b^2}=\frac{\tan B}{\tan C}$.