Proving Trig Identities Prove the following using trigonometric identities. \( \begin{array}{lll}\text { 44. } \frac{\tan ^{2} x}{\sin ^{2} x}=\frac{1}{\cos ^{2} x} & \text { 45. } \sec ^{2} x+\csc ^{2} x=\frac{\csc ^{2} x}{\cos ^{2} x}\end{array} \)

To prove that \( \frac{\tan^2 x}{\sin^2 x} = \frac{1}{\cos^2 x} \), start by recalling that \( \tan x = \frac{\sin x}{\cos x} \). Therefore, \( \tan^2 x = \frac{\sin^2 x}{\cos^2 x} \). Now substituting this into the left side gives us: \[ \frac{\tan^2 x}{\sin^2 x} = \frac{\frac{\sin^2 x}{\cos^2 x}}{\sin^2 x} = \frac{1}{\cos^2 x} \] This proves the first identity. For the second identity, \( \sec^2 x + \csc^2 x = \frac{\csc^2 x}{\cos^2 x} \), start by recalling \( \sec^2 x = \frac{1}{\cos^2 x} \) and \( \csc^2 x = \frac{1}{\sin^2 x} \). Therefore, the left side becomes: \[ \sec^2 x + \csc^2 x = \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} \] To combine these fractions, find a common denominator, which is \( \sin^2 x \cos^2 x \): \[ \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} \] Since \( \sin^2 x + \cos^2 x = 1 \), the expression simplifies to: \[ \frac{1}{\sin^2 x \cos^2 x} \] Now, note that \( \frac{1}{\sin^2 x \cos^2 x} = \frac{\csc^2 x}{\cos^2 x} \) proves our identity. Thus, both identities are verified!