<s>[INST] 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
 [/INST] Multiplying  the numerator & the denominator by $\cos^2x,$
$$\frac{2+\tan^2x}{\sec^2x}-1=2\cos^2x+\sin^2x-1$$
$$=2\cos^2x-(1-\sin^2x)=\cos^2x$$
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