Answer :
Use the trigonometric identities:
[tex]\tan \theta = \frac{\sin \theta}{\cos \theta} \\ \sin^2 \theta+ \cos^2 \theta=1[/tex]
Proof:
[tex]\frac{\tan^2 \theta}{1+ \tan^2 \theta}}=\sin^2 \theta \ \ \ |\times (1+\tan^2 \theta) \\ \\ \tan^2 \theta = \sin^2 \theta (1+ \tan^2 \theta) \ \ \ |\hbox{convert } \tan \theta \hbox{ to } \frac{\sin \theta}{\cos \theta} \\ \\ (\frac{\sin \theta}{\cos \theta})^2=\sin^2 \theta (1+ (\frac{\sin \theta}{\cos \theta})^2) \\ \\ \frac{\sin^2 \theta}{\cos^2 \theta}=\sin^2 \theta(1+\frac{\sin^2 \theta}{\cos^2 \theta}) \ \ \ |\div \sin \theta[/tex]
[tex]\frac{1}{\cos^2 \theta}=1+\frac{sin^2 \theta}{\cos^2 \theta} \ \ \ |\times \cos^2 \theta \\ \\ 1=\cos^2 \theta+\sin^2 \theta \ \ \ |\hbox{convert } \cos^2 \theta+ \sin^2 \theta \hbox{ to } 1 \\ \\ 1=1[/tex]
[tex]\tan \theta = \frac{\sin \theta}{\cos \theta} \\ \sin^2 \theta+ \cos^2 \theta=1[/tex]
Proof:
[tex]\frac{\tan^2 \theta}{1+ \tan^2 \theta}}=\sin^2 \theta \ \ \ |\times (1+\tan^2 \theta) \\ \\ \tan^2 \theta = \sin^2 \theta (1+ \tan^2 \theta) \ \ \ |\hbox{convert } \tan \theta \hbox{ to } \frac{\sin \theta}{\cos \theta} \\ \\ (\frac{\sin \theta}{\cos \theta})^2=\sin^2 \theta (1+ (\frac{\sin \theta}{\cos \theta})^2) \\ \\ \frac{\sin^2 \theta}{\cos^2 \theta}=\sin^2 \theta(1+\frac{\sin^2 \theta}{\cos^2 \theta}) \ \ \ |\div \sin \theta[/tex]
[tex]\frac{1}{\cos^2 \theta}=1+\frac{sin^2 \theta}{\cos^2 \theta} \ \ \ |\times \cos^2 \theta \\ \\ 1=\cos^2 \theta+\sin^2 \theta \ \ \ |\hbox{convert } \cos^2 \theta+ \sin^2 \theta \hbox{ to } 1 \\ \\ 1=1[/tex]
