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Alex787 [66]
3 years ago
7

If sin ∅ = m-n/m+n, find the value of1 + Tan²∅​

Mathematics
2 answers:
Vilka [71]3 years ago
3 0

Answer:

l hope it will help you thanks

kow [346]3 years ago
3 0

Answer:

1+\tan^2\theta = \frac{(m+n)^2}{4mn}

Step-by-step explanation:

We will use the identity:

                                      1+\tan^2\theta = \sec^2\theta

Also,

                                          \sec^2\theta = \dfrac{1}{\cos^2\theta}

Therefore,

                                      1+\tan^2\theta = \dfrac{1}{\cos^2\theta}

Now we will use the identitiy:

                                      \sin ^2 \theta + \cos ^2 \theta = 1

Therefore we obtain:

                                      \cos ^2 \theta = 1-\sin^2 \theta

Inserting \sin ∅ =  (m-n)/(m+n) into the above equation we obtain:

                                     \cos^2\theta = 1 - \dfrac{(m-n)^2}{(m+n)^2}

Editing that expression:

       \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}

               = \dfrac{m^2+2mn+n^2 - m^2+2mn-n^2}{(m+n)^2}=\dfrac{4mn}{(m+n)^2}

Now we have the expression for \cos^2\theta, and since   1+\tan^2\theta = \frac{1}{\cos^2\theta}  

we need  \frac{1}{\cos^2\theta}.

Therefore:

                                       1+\tan^2\theta = \frac{(m+n)^2}{4mn}

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