Question

If tanA +sinA=P and tanA-sinA=Q to prove P^2-Q^2=4√PQ​

Answer 1

$tana + sina = p \\ \implies( {tana + sina)}^{2} = {p}^{2} ...(1) \\ \\ tana - sina = q \\ \implies( {tana - sina)}^{2} = {q}^{2} ...(2) \\ \\ (1) - (2) \\ {p}^{2} - {q}^{2} = ( {tan a+ sina})^{2} - ( {tana - sina)}^{2} \\ {p}^{2} - {q}^{2} = 4 \: tanasina \\ {p}^{2} - {q}^{2} = \sqrt{ {tan}^{2} a {sin}^{2} a} \\ {p}^{2} - {q}^{2} = 4 \sqrt{ {tan}^{2} a(1 - {cos}^{2} a)} \\ {p}^{2} - {q}^{2} = 4 \sqrt{ {tan}^{2}a - {tan}^{2} a {cos}^{2} a } \\ {p}^{2} - {q}^{2} = 4 \sqrt{ {tan}^{2}a - {sin}^{2} a} \\ {p}^{2} - {q}^{2} = 4 \sqrt{(tana + sina)(tana - sina)} \\ {p}^{2} - {q}^{2} = 4 \sqrt{pq}$