Question

tan theta + sin theta = m and tan theta - sin theta = n.

Prove that m^2 - n^2 = 4root (mn).

Answer 1

$\textbf{ \underline{ \underline{Question - }}}$

$tan \theta + sin \theta = m \\ tan \theta \: - sin \theta = n$
Prove that
${m}^{2} - {n}^{2} = 4 \sqrt{mn}$

$\textbf{ \underline{ \underline{Solution - }}}$

$\textsf{ \underline{ \underline{step : 1}}}$

$\text{Find the value of mn}$
$mn = (tan \theta + sin \theta)(tan \theta - sin \theta) \\ mn = {tan}^{2} \theta - {sin}^{2} \theta \\ mn = \frac{ {sin}^{2} \theta}{ {cos}^{2} \theta} - \frac{ {sin}^{2} \theta}{1} \\ mn = {sin}^{2} \theta( \frac{1}{ {cos}^{2} \theta} - \frac{1}{1} ) \\ mn = {sin}^{2} \theta( \frac{1 - {cos}^{2} \theta }{ {cos}^{2} \theta } ) \\ mn = {sin}^{2} \theta \times \frac{ {sin}^{2} \theta}{ {cos}^{2} \theta } \\ mn = {sin}^{2} \theta \times {tan}^{2} \theta$


$\textsf {\underline {\underline {step:2}}}$

The value of

$\sqrt{mn} = \sqrt{ {sin}^{2} \theta \times {tan}^{2} \theta} \\ \sqrt{mn} = sin \theta \: tan \theta$

${m}^{2} - {n}^{2} = 4 \sqrt{mn } \\ (tan \theta + sin \theta)^{2} - (tan \theta - sin \theta)^{2} = 4sin \theta \: tan \theta$

$\text{Taking LHS}$

$(tan \theta + sin \theta)^{2} - (tan \theta - sin \theta)^{2} \\ = >( {tan}^{2} \theta + {sin}^{2} \theta + 2tan \theta \: sin \theta) - ( {tan}^{2} \theta + {sin}^{2} \theta - 2tan \theta \: sin \theta) \\ = > {tan}^{2} \theta + {sin}^{2} \theta + 2tan \theta \: sin \theta - {tan}^{2} \theta - {sin}^{2} \theta + 2tan \theta \: sin \theta \\ = > 4tan \theta \: sin \theta$

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

$\text{ LHS = RHS }$

$\textbf{Hence Proved}$

Answer 2

solution is attached