# The area of triangle ABC is equal to a^2+b^2-c^2. If angle C is acute then find the numerical vaue of secant where a,b,c are positive real numbers?

Jul 26, 2018

Given $\Delta = {a}^{2} + {b}^{2} - {c}^{2.} . . \left(1\right)$

We know

$\cos C = \frac{{a}^{2} + {b}^{2} - {c}^{2}}{2 a b} \ldots . \left(2\right)$

$\Delta = \frac{1}{2} a b \sin C$

$\implies a b = \frac{2 \Delta}{\sin} c \ldots . \left(3\right)$

Combining these equtions we get

$\cos C = \frac{\Delta \sin C}{2 \cdot 2 \Delta}$

$\implies \tan C = 4$

$\implies {\sec}^{2} C = 1 + {\tan}^{2} C = 1 + 16 = 17$

$\implies \sec C = \sqrt{17}$