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Prove this identity. Tan^2(0)/1 + tan^2(0) = sin^2(0). Note i placed a bracket around the zeros because i didnt want to look like it is to the power of 20. And the zero is supposed to be the symbol theta.



Answer :

naǫ
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]

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