Question

If sinA + cosA = V2cosA, (0 ≠90°) then the value of tanA is​

Answer 1

$sinx + cosx = \sqrt{2} cosx \\ sinx = \sqrt{2} cosx - cosx \\ sinx = cosx( \sqrt{2} - 1) \\ \\ squaring \: on \: both \: sides \\ \\ {sin}^{2} x = {cos}^{2} (2 + 1 - 2 \sqrt{2} ) \\ {sin}^{2} x = (1 - {sin}^{2} x)(3 - 2 \sqrt{2} ) \: \: \: ...( {sin}^{2} x + {cos}^{2} x = 1) \\ {sin}^{2} x = 3 - 2 \sqrt{2} - 3 {sin}^{2} x + 2 \sqrt{2} {sin}^{2} x \\ {sin}^{2} x + 3 {sin}^{2} x - 2 \sqrt{2} {sin}^{2} x + 2 \sqrt{2} - 3 = 0 \\ 4 {sin}^{2} x - 2 \sqrt{2} {sin}^{2} x + 2 \sqrt{2} - 3 = 0 \\ {sin}^{2} x(4 - 2 \sqrt{2} ) = 3 - 2 \sqrt{2 } \\ {sin}^{2} x = \frac{3 - 2 \sqrt{2} }{4 - 2 \sqrt{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(i) \\ 1 - {cos}^{2} x = \frac{3 - 2 \sqrt{2} }{4 - 2 \sqrt{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: ... \: ( {sin}^{2} x + {cos}^{2} x = 1) \\ {cos}^{2} x = 1 - \frac{(3 - 2 \sqrt{2}) }{4 - 2 \sqrt{2} } = \frac{4 - 2 \sqrt{2} - 3 + 2 \sqrt{2} }{4 - 2 \sqrt{2} } \\ {cos}^{2} x = \frac{1}{4 - 2 \sqrt{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ... \: (ii) \\ \\ dividing \: (i) \: by \: (ii) \\ \\ \frac{ {sin}^{2}x }{ {cos}^{2}x } = \frac{ \frac{3 -2 \sqrt{2} }{4 - 2 \sqrt{2} } }{ \frac{1}{4 - 2 \sqrt{2} } } = ( \frac{3 - 2 \sqrt{2} }{4 - 2 \sqrt{2} } )( \frac{4 - 2 \sqrt{2} }{1} ) \\ {tan}^{2} x = 3 - 2 \sqrt{2} \: \: \: \: \: \: \: \: \: \: ... \: (tanx = \frac{sinx}{cos} ) \\ \\ taking \: square \: roots \: on \: both \: sides \\ \\ tanx = \sqrt{3 - 2 \sqrt{2} }$