Respuesta :
Let's rewrite the identity in two different forms conserving the functions given:
[tex]\tan ^2\theta-\sec ^2\theta=-1[/tex]and
[tex]\tan ^2\theta=\sec ^2\theta-1[/tex]Now, let's find two different identities from this one. To do this let's remember that:
[tex]\begin{gathered} \tan \theta=\frac{\sin \theta}{\cos \theta} \\ \sec \theta=\frac{1}{\cos \theta} \end{gathered}[/tex]Then we have:
[tex]\begin{gathered} \tan ^2\theta+1=\sec ^2\theta \\ \tan ^2\theta-\sec ^2\theta=-1 \\ \frac{\sin^2\theta}{\cos^2\theta}-\frac{1}{\cos^2\theta}=-1 \\ \frac{\sin^2\theta-1}{\cos^2\theta}=-1 \\ \sin ^2\theta-1=-\cos ^2\theta \\ \sin ^2\theta+\cos ^2\theta=1 \end{gathered}[/tex]Therefore, we can write the identity given as:
[tex]\sin ^2\theta+\cos ^2\theta=1[/tex]Let's find a second identity from the one given, to do this we will mutiply it by the cotangent squared of the angle:
[tex]\begin{gathered} \tan ^2\theta+1=\sec ^2\theta \\ \tan ^2\theta\cdot\cot ^2\theta+\cot ^2\theta=\sec ^2\theta\cdot\cot ^2\theta \\ \frac{\sin^2\theta}{\cos^2\theta}\cdot\frac{\cos^2\theta}{\sin^2\theta}+\cot ^2\theta=\frac{1}{\cos^2\theta}\cdot\frac{\cos ^2\theta}{\sin ^2\theta} \\ 1+\cot ^2\theta=\frac{1}{\sin ^2\theta} \\ 1+\cot ^2\theta=\csc ^2\theta \end{gathered}[/tex]Therefore, we can write the identity given as:
[tex]1+\cot ^2\theta=\csc ^2\theta[/tex]