Question

if tanθ + sinθ = m and tanθ - sinθ = n prove that (m^2 - n^2) = 4√mn ​

Answer 1

Given that,

$\rm tan\theta + sin \theta = m$

$\rm tan\theta - sin\theta = n$

To prove,

$\rm(m^2 - n^2) = 4 \sqrt{mn}$

Proof,

$\rm LHS = (m^2 - n^2)$

$\rm = (tan\theta + sin \theta)^2 - (tan\theta - sin \theta)^2$

we know that,

(a + b)² - (a - b)² = 4ab

$\rm = 4 \: tan\theta \: sin \theta$

Now,

$\rm \: RHS = 4 \sqrt{mn}$

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

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

$\rm = 4 \sqrt{ \bigg( \dfrac{ {sin}^{2} \theta }{ {cos}^{2} \theta } - sin \theta^{2} \bigg)}$

$\rm = 4 \sqrt{ \bigg( \dfrac{ {sin}^{2} \theta - sin \theta^{2} {cos}^{2} \theta }{ {cos}^{2} \theta } \bigg)}$

$\rm = 4 \sqrt{ \dfrac{ {sin}^{2} \theta(1 - {cos}^{2} \theta )}{ {cos}^{2} \theta}}$

$\rm = 4 \: \dfrac{sin\theta}{ cos\theta} \sqrt{(1 - {cos}^{2} \theta )}$

$\rm = 4 \: tan\theta \: sin \theta$

Hence, LHS = RHS

so,

$\rm(m^2 - n^2) = 4 \sqrt{mn}$

Hence proved !!