Lost solving a trigonometric identity $$\frac { 2+\tan^2x }{  \sec^2x} -1$$
Steps I took:
$$\begin{align}
\frac { 2+\tan^2x }{  \sec^2x} -1
&=\frac { 2+\frac { \sin^{ 2 }x }{ \cos^{ 2 }x }  }{ \frac { 1 }{ \cos^2x }  } -1\\
&=\frac { 2\cos^{ 2 }x+\sin^{ 2 }x }{ \cos^{ 2 }x } \cdot \frac { \cos^{ 2 }x }{ 1 } -1\\
&=\frac { (2\cos^{ 2 }x+\sin^{ 2 }x)\cos^{ 2 }x }{ \cos^{ 2 }x } -1\\
&=2\cos^{ 2 }x+\sin^{ 2 }x-1\\
\end{align}$$
Now I have no idea where to go from here.. Please guide me in the right direction

$$
\frac{2+\tan^2 x}{\sec^2 x} - 1= \frac{1 + 1 +\tan^2 x}{\sec^2 x} - 1 = \frac{1 + \sec^2 x}{\sec^2 x} - 1.
$$
thus you end up getting
$$
\frac{1}{\sec^2 x} + 1 - 1 = \cos^2 x
$$
dunno how this helps you in the long run..
