# What is the slope of the tangent line of sin^2y-cos^2x= C , where C is an arbitrary constant, at (pi/3,pi/3)?

Jan 20, 2018

Slope is $1$

#### Explanation:

As we are seeking slope at $\left(\frac{\pi}{3} , \frac{\pi}{3}\right)$, this point lies on the curve given by ${\sin}^{2} y - {\cos}^{2} x = C$.

Putting values from $\left(\frac{\pi}{3} , \frac{\pi}{3}\right)$, we get $\frac{3}{4} - \frac{1}{4} = C$

or $C = \frac{1}{2}$ and function is ${\sin}^{2} y - {\cos}^{2} x = \frac{1}{2}$

Now differentiating the implicit function ${\sin}^{2} y - {\cos}^{2} x = \frac{1}{2}$

$2 \sin y \cos y \frac{\mathrm{dy}}{\mathrm{dx}} + 2 \cos x \sin x = 0$

or $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 \sin x \cos x}{2 \sin y \cos y} = - \sin \frac{2 x}{\sin} \left(2 y\right)$

and at $\left(\frac{\pi}{3} , \frac{\pi}{3}\right)$, slope is $\frac{\frac{1}{2}}{\frac{1}{2}} = 1$