Given:
Equation of Curve is:
[tex]y=\frac{1}{2}\ln (\sin ^2x)[/tex]
To find: Equation of tangent to the curve at point
[tex](\frac{\pi}{4},\frac{-1\ln 2}{2})[/tex]
Equation of tangent to the curve is given by:
[tex]y-y_1=m(x-x_1)[/tex]
where, m is slope of line.
Now, m is derivative of y with respect to x at given point.
Hence,
[tex]\begin{gathered} m=\frac{1}{2}\text{ }\times\frac{2\sin x\cos x}{\sin^2x} \\ m=\frac{\cos \text{ x}}{\sin \text{ x}} \\ m=\cot \text{ x} \end{gathered}[/tex]
At given point, the slope m is:
[tex]\begin{gathered} m=\cot (\frac{\pi}{4}) \\ m=1 \end{gathered}[/tex]
Therefore,the equation of tangent to the curve is given as:
[tex]\begin{gathered} y-y_1=1(x-x_1) \\ y-(\frac{-1(\ln \text{ 2))}}{2})=x-\frac{\pi}{4} \\ \frac{2y+\ln \text{ 2}}{2}=\frac{4x-\pi}{4} \\ 2y+\ln 2=\frac{4x-\pi}{2} \end{gathered}[/tex][tex]\begin{gathered} 4y+2\ln 2=4x-\pi \\ 4y=4x-\pi-2\ln 2 \end{gathered}[/tex]
Thus the required equation of tangent to the curve is
[tex]4y=4x-\pi-2\ln 2[/tex]
