Answer:
[tex]1+\tan^2\theta = \frac{(m+n)^2}{4mn}[/tex]
Step-by-step explanation:
We will use the identity:
[tex]1+\tan^2\theta = \sec^2\theta[/tex]
Also,
[tex]\sec^2\theta = \dfrac{1}{\cos^2\theta}[/tex]
Therefore,
[tex]1+\tan^2\theta = \dfrac{1}{\cos^2\theta}[/tex]
Now we will use the identitiy:
[tex]\sin ^2 \theta + \cos ^2 \theta = 1[/tex]
Therefore we obtain:
[tex]\cos ^2 \theta = 1-\sin^2 \theta[/tex]
Inserting \sin ∅ = (m-n)/(m+n) into the above equation we obtain:
[tex]\cos^2\theta = 1 - \dfrac{(m-n)^2}{(m+n)^2}[/tex]
Editing that expression:
[tex]\cos ^2 \theta = \dfrac{(m+n)^2-(m-n)^2}{(m+n)^2} = \dfrac{m^2+2mn+n^2 - (m^2-2mn+n^2)}{(m+n)^2}[/tex]
[tex]= \dfrac{m^2+2mn+n^2 - m^2+2mn-n^2}{(m+n)^2}=\dfrac{4mn}{(m+n)^2}[/tex]
Now we have the expression for [tex]\cos^2\theta[/tex], and since [tex]1+\tan^2\theta = \frac{1}{\cos^2\theta}[/tex]
we need [tex]\frac{1}{\cos^2\theta}[/tex].
Therefore:
[tex]1+\tan^2\theta = \frac{(m+n)^2}{4mn}[/tex]
