Trigonometric Functions — Maths STD 12 Science — Question

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

$\therefore a=f k \sin A, b=k \sin B, c=k \sin C$

Now, $\left(c^2-a^2+b^2\right) \tan \mathrm{A}=\left(c^2-a^2+b^2\right) \cdot \frac{\sin \mathrm{A}}{\cos \mathrm{A}}$

$=\left(c^2+b^2-a^2\right) \times \frac{k a}{\left(\frac{c^2+b^2-a^2}{2 b c}\right)}$

$=\left(c^2+b^2-a^2\right) \times \frac{2 k a b c}{c^2+b^2-a^2}$

$=2 k a b c$

$\ldots(1)$

$\left(a^2-b^2+c^2\right) \tan \mathrm{B}=\left(a^2-b^2+c^2\right) \cdot \frac{\sin \mathrm{B}}{\cos \mathrm{B}}$

$=\left(a^2+c^2-b^2\right) \times \frac{k b}{\left(\frac{a^2+c^2-b^2}{2 a c}\right)}$

$=\left(a^2+c^2-b^2\right) \times \frac{2 k a b c}{a^2+c^2-b^2}$

$=2 \mathrm{kabc}$

$\ldots(2)$

$\left(b^2-c^2+a^2\right) \tan C=\left(b^2-c^2+a^2\right) \cdot \frac{\sin C}{\cos C}$

$=\left(a^2+b^2-c^2\right) \times \frac{k c}{\left(\frac{a^2+b^2-c^2}{2 a b}\right)}$

$\begin{aligned} & =\left(a^2+b^2-c^2\right) \times \frac{2 k a b c}{a^2+b^2-c^2} \\ & =2 k a b c\end{aligned}$

... (3)From (1), (2) and (3), we get

$\begin{aligned} & \left(c^2-a^2+b^2\right) \tan A=\left(a^2-b^2+c^2\right) \tan B \\ & =\left(b^2-c^2+a^2\right) \tan C .\end{aligned}$