Question

if tanA + sinA =m and tanA - sinA = n then m^2 - n^2 =4 route over mn​

Answer 1

I hope it will help you.

Answer 2

$\huge{\mathfrak{Question:-}}$

$\bold{tanA + sinA = m \; and\; tanA-sinA = n, \; then\; m^{2}-n^{2} =4\sqrt{mn}}$

$\huge{\mathfrak{Answer:-}}$

$\large{\textbf{\underline{\underline{Given:-}}}}$

$\bold{tanA+sinA = m \;\;\;\;\;...........(1)}\\ \\ \\\bold{tanA-sinA = n \;\;\;\;\;...........(2)}$

Adding equation(1) and (2) we get,

$\bold{2\;tanA= m+n}\\ \\ \\ \bold{\implies tanA = \frac{m+n}{2}}$

$\bold{Subtituting\;tanA = \frac{m+n}{2}\; in\;equation\;(1),}$

$\bold{\frac{m+n}{2}+sinA = m}$

$\bold{\implies sinA=m-\frac{m+n}{2} = \frac{2m-m-n}{2} = \frac{m-n}{2}}$

$\bold{It\;is\;known\;that\;sec^{2}A-tan^{2}A=1}$

$\bold{\implies \frac{1}{cos^{2}A}-tan^{2}A =1}$

$\bold{\implies \frac{1}{1-sin^{2}A}-tan^{2} A = 1}$

$\bold{\implies \frac{1-(\frac{m+n}{2})^{2}+(\frac{m+n}{2})^{2}+(\frac{m-n}{2})^{2}}{1-(\frac{m-n}{2})^{2}} = 1}$

$\bold{\implies 1-(\frac{m+n}{2})^{2}+[\frac{(m-n)(m+n)}{4}]^{2} = 1-(\frac{m-n}{2})^{2}}$

$\bold{\implies (\frac{m^{2}-n^{2}}{4})^2 = (\frac{m+n}{2})^{2}-(\frac{m-n}{2})^{2}}$

$\bold{= \frac{m^{2}+n^{2}+2mn}{4} -\frac{m^{2}+n^{2}-2mn}{4}}\\ \\ \\ \bold{= \frac{m^{2}+n^{2}+2mn-m^{2}-n^{2}+2mn}{4}}$

$\bold{\implies (\frac{m^{2}-n^{2}}{4})^{2} = \frac{4\,mn}{4} = mn}\\ \\ \\ \implies \bold{\frac{m^{2}-n^{2}}{4} =\sqrt{mn}} \\ \\ \\\bold{\implies m^{2}-n^{2} = 4\sqrt{mn}}$