Question

I need help in math can you please help me

Answer 1

To verify the identity we need to remember the definitions of the trigonometric functions:

$\begin{gathered} \tan \theta=\frac{\sin \theta}{\cos \theta} \\ \cot \theta=\frac{\cos \theta}{\sin \theta} \\ \sec \theta=\frac{1}{\cos \theta} \\ \csc \theta=\frac{1}{\sin \theta} \end{gathered}$

With this in mind:

$\begin{gathered} \text{tan}^2\theta-\cot ^2\theta=\frac{\sin^2\theta}{\cos^2\theta}-\frac{\cos ^2\theta}{\sin ^2\theta} \\ =\frac{\sin ^4\theta-\cos ^4\theta}{\cos ^2\theta\sin ^2\theta} \\ =\frac{(\sin ^2\theta+\cos ^2\theta)(\sin ^2\theta-\cos ^2\theta)}{\cos ^2\theta\sin ^2\theta} \end{gathered}$

Now we need to remembert the identity:

$\sin ^2\theta+\cos ^2\theta=1$

then:

$\begin{gathered} \tan ^2\theta-\cot ^2\theta=\frac{(\sin^2\theta+\cos^2\theta)(\sin^2\theta-\cos^2\theta)}{\cos^2\theta\sin^2\theta} \\ =\frac{\sin ^2\theta-\cos ^2\theta}{\cos ^2\theta\sin ^2\theta} \\ =\frac{\sin^2\theta}{\cos^2\theta\sin^2\theta}-\frac{\cos ^2\theta}{\cos ^2\theta\sin ^2\theta} \\ =\frac{1}{\cos^2\theta}-\frac{1}{\sin ^2\theta} \\ =\sec ^2\theta-\csc ^2\theta \end{gathered}$