Question

TanA+sinA=m and tanA-sinA=n then prove that m^2-n^2=√mn

Answer 1

Answer:

I don't know this answer sorry I can't understand

Answer 2

Answer:

Proved below.

Step-by-step explanation:

tanA + sinA = m -----> (i)

tanA - sinA = n --------->(ii)

$m^{2} - n^{2} = (m - n)(m + n)\\m^{2} - n^{2} = ( tanA + sinA - tanA +sinA)( tanA + sinA + tanA - sinA)\\m^{2} - n^{2} = 2sinA . 2tanA\\m^{2} - n^{2} = 4.sinA.tanA$

$4\sqrt{mn} = 4\sqrt{(tanA + sinA)(tanA - sinA)}\\4\sqrt{mn} =4 \sqrt{tanA^{2} - sinA^{2}} \\4\sqrt{mn} = 4 \sqrt{\frac{sinA^{2} - sinA^{2}cosA^{2} }{cosA^{2}} }\\4\sqrt{mn} = 4 \sqrt{\frac{sinA^{2}(1 - cosA^{2} ) }{cosA^{2} } }\\\\4\sqrt{mn} = 4\sqrt{\frac{sinA^{2} }{cosA^{2} } . sinA^{2} }\\ 4\sqrt{mn} = 4 \sqrt{tanA^{2}.sinA^{2} }\\ 4\sqrt{mn} = 4.tanA.sinA$

From above, it is proved