Question

If sec +tan=m, then prove that sin =m2-1/m2+1​

Answer 1

$secx + tanx = m \: \: \: \: \: \: \: \: \: ... \: (i) \: ... \: (given)\\ \\ we \: know \: that \\ 1 + {tan}^{2} x = {sec}^{2} x \\ {sec}^{2}x - {tan}^{2} x = 1 \\ (secx + tanx)(secx - tanx) = 1 \\ m(secx - tanx) = 1 \\ secx - tanx = \frac{1}{m} \: \: \: \: \: \: \: \: ... \: (ii) \\ \\ adding \: (i) \: and \: (ii) \\ \\ secx + tanx + secx - tanx = m + \frac{1}{m} \\ 2secx = \frac{ {m}^{2} + 1}{m} \\ secx = \frac{ {m}^{2} + 1 }{2m} \\ cosx = \frac{2m}{ {m}^{2} + 1} \: \: \: \: \: \: \: \: \: \: \: ... \: (cosx = \frac{1}{secx} ) \\ {cos}^{2} x = \frac{ {4m}^{2} }{ {m}^{4} + 1 + 2 {m}^{2} } \\ \\ now \\ we \: know \: that \\ \\ {sin}^{2} x + {cos}^{2} x = 1 \\ {sin}^{2} x = 1 - {cos}^{2} x \\ {sin}^{2} x = 1 - \frac{ {4m}^{2} }{ {m}^{4} + 1 + 2 {m}^{2} } \\ {sin}^{2} x = \frac{ {m}^{4} + 1 + 2 {m}^{2} - 4 {m}^{2} }{ {( {m}^{2} + 1)}^{2} } \\ {sin}^{2} x = \frac{ {m}^{4} + 1 - 2 {m}^{2} }{ {( {m}^{2} + 1) }^{2} } \\ {sin}^{2} x = \frac{ {( {m}^{2} - 1) }^{2} }{ {( {m}^{2} + 1)}^{2} } \\ \\ taking \: square \: root \: on \: both \: sides \\ \\ sinx = \frac{ {m}^{2} - 1}{ {m}^{2} + 1 }$

... Hence Proved!