# If (tan θ + sin θ) = m and (tan θ − sin θ) = n,

Question:

If (tan θ + sin θ) = m and (tan θ − sin θ) = n, prove that (m2 − n2)2 = 16mn.

Solution:

We have $(\tan \theta+\sin \theta)=m$ and $(\tan \theta-\sin \theta)=n$

Now, LHS $=\left(m^{2}-n^{2}\right)^{2}$

$=\left[(\tan \theta+\sin \theta)^{2}-(\tan \theta-\sin \theta)^{2}\right]^{2}$

$=\left[\left(\tan ^{2} \theta+\sin ^{2} \theta+2 \tan \theta \sin \theta\right)-\left(\tan ^{2} \theta+\sin ^{2} \theta-2 \tan \theta \sin \theta\right)\right]^{2}$

$=\left[\left(\tan ^{2} \theta+\sin ^{2} \theta+2 \tan \theta \sin \theta-\tan ^{2} \theta-\sin ^{2} \theta+2 \tan \theta \sin \theta\right)\right]^{2}$

$=(4 \tan \theta \sin \theta)^{2}$

$=16 \tan ^{2} \theta \sin ^{2} \theta$

$=16 \frac{\sin ^{2} \theta}{\cos ^{2} \theta} \sin ^{2} \theta$

$=16 \frac{\left(1-\cos ^{2} \theta\right) \sin ^{2} \theta}{\cos ^{2} \theta}$

$=16\left[\tan ^{2} \theta\left(1-\cos ^{2} \theta\right)\right]$

$=16\left(\tan ^{2} \theta-\tan ^{2} \theta \cos ^{2} \theta\right)$

$=16\left(\tan ^{2} \theta-\frac{\sin ^{2} \theta}{\cos ^{2} \theta} \times \cos ^{2} \theta\right)$

$=16\left(\tan ^{2} \theta-\sin ^{2} \theta\right)$

$=16(\tan \theta+\sin \theta)(\tan \theta-\sin \theta)$

$=16 m n \quad[(\tan \theta+\sin \theta)(\tan \theta-\sin \theta)=m n]$

$\therefore\left(m^{2}-n^{2}\right)\left(m^{2}-n^{2}\right)^{2}=16 m n$