Question

If tan a +sin a =m and tan a -sin a =n then show m^2-n^2=4√mn

Answer 1

$m = tana + sina \\ m {}^{2} = {tan}^{2} a + {sin}^{2} a + 2tana \: sina \\ n {}^{2} = {tan}^{2} a + {sin}^{2} a - 2tanasina \\ subtract \\ {m}^{2} - {n}^{2} = 4tanasina.\\ {m}^{2} - {n}^{2} = 4 {sin}^{2} a \div cosa ....(1) \\ now \: mn = {tan}^{2} a - {sin}^{2} a = {sin}^{4}a \div {cos}^{2} a \\ \sqrt{mn} = {sin}^{2} a \div cosa \\ put \: in \: (1) \\ {m}^{2} - {n}^{2} = 4 \sqrt{mn} \\ proved \\ mark \: it \: brainliest \: if \: helped \\ ..........$
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