Question

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

Answer 1

Answer:

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Answer 2

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:

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