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Q.
The area of a triangle $A B C$ is equal to $\left(a^{2}+b^{2}-c^{2}\right)$, where $a, b$ and $c$ are the sides of the triangle. The value of $\tan C$ equals
Trigonometric Functions
Solution:
$\cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}$
Now, $\Delta=a^{2}+b^{2}-c^{2}$
Hence, $\cos C=\frac{\Delta}{2 a b}$ ...(1)
Also, $\Delta=\frac{1}{2} a b \sin C$
$\Rightarrow \frac{2 \Delta}{\sin C}=a b$
$\Rightarrow \sin C=\frac{2 \Delta}{a b}$ ...(2)
From (1) and (2), we get
$\tan C=\frac{2 \Delta}{a b} \cdot \frac{2 a b}{\Delta}=4$