Question

plz answer this it's argent ​

Answer 1

Answer:

Proved: $m^{2}-n^{2}=4\sqrt{mn}$

Step-by-step explanation:

It is provided that:

$tan\theta+sin\theta=m\\tan\theta-sin\theta=n$

Solve the left hand side of the provided equation to compute the right hand side as follows:

$m^{2}-n^{2}=(tan\theta+sin\theta)^{2}-(tan\theta-sin\theta)^{2}\\=tan^{2}\theta+sin^{2}\theta+2tan\theta sin\theta-tan^{2}\theta-sin^{2}\theta+2tan\theta sin\theta\\=4tan\theta sin\theta\\=4\times tan\theta \sqrt{1-cos^{2}\theta}\\=4\times \sqrt{tan^{2}\theta(1-cos^{2}\theta)}\\=4\times\sqrt{tan^{2}\theta-tan^{2}\theta cos^{2}\theta}\\=4\times\sqrt{tan^{2}\theta-\frac{sin^{2}\theta}{cos^{2}\theta}\times cos^{2}\theta}\\=4\times\sqrt{tan^{2}\theta-sin^{2}\theta}$

$=4\times\sqrt{(tan\theta+sin\theta)(tan\theta-sin\theta)}\\=4\sqrt{mn}$

Hence proved.