Question 3

If $$\tan \theta + \sin \theta = m$$ and $$\tan \theta - \sin \theta = n$$, then the value of $$m^2 - n^2$$ is equal to

Solution

$$m^2-n^2=\ \left(m+n\right)\left(m-n\right)=\left(2\tan\theta\ \right)\left(2\sin\theta\ \right)=4\tan\theta\ \sin\theta\ $$

Going by the options you have only mn or m/n

Case 1 : mn

mn= $$\left(\tan\theta\ +\sin\theta\right)\ \left(\tan\theta\ -\sin\theta\ \right)$$ = $$\tan^2\theta\ -\sin^2\theta$$ = $$\frac{\sin^2\theta}{\cos^2\theta\ }\ -\sin^2\theta\ =\sin^2\theta\ \left(\frac{1}{\cos^2\theta\ }-1\right)=\frac{\sin^2\theta}{\cos^2\theta\ }\ \left(1-\cos^2\theta\ \right)=\tan^{^2}\theta\ \cdot\sin^2\theta\ $$

So $$4\sqrt{\ mn}\ =\ 4\tan\theta\ \cdot\sin\theta\ $$


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