If \sin \theta = \frac{m^2-n^2}{m^2+n^2}, then what is the value of tan θ?

Trigonometry Practice Test 2

Question : 2 of 22

Marks: +1, -0

\sin θ = {m^2-n^2}/{m^2+n^2}

, then what is the value of tan θ?

Solution:

In ∆ABC,

AB=√{(AC)^2-(BC)^2}

=√{m^4+n^4+2m^2n^2-(m^4+n^4-2m^2n^2)}

=√{4m^2n^2}=2mn

∴\tan θ={m^2-n^2}/{2mn}

Go to Question:

Prev Question

Next Question