Question

If $\frac{1-tanx}{1+tany}$ = tany and x-y=pi/4 what is x?
a) (1-2n)$\frac{\pi}{4}$
b) (1+2n)$\frac{\pi}{4}$
c) (2n-1)$\frac{\pi}{3}$
d) (2n+1)$\frac{\pi}{6}$

Answer 1

Answer:

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Step-by-step explanation:

$to \: find = \\ value \: of \: x \\ \\ so \: here \: \\ x - y = \frac{\pi}{4} \\ \\ ie \: \: x - y = 45 \\ so \: \: x = y + 45 \: \: \: \: \: \: \: \: \: \: \: (1)$

$here \\ \frac{1 - tan.x}{1 + tan.y} = tan.y \\ \\ 1 - tan.x = tan.y \: (1 + tan.y) \\ 1 - tan.x = tan.y + tan {}^{2} .y \\ \\ 1 - tan \: (y + 45) = tan.y + tan {}^{2} .y \\ so \: here \: tan \: (y + 45) \: is \: in \: form \: tan \: (A+B) \\ \\ so \: we \: know \: that \\ tan \: (A+B) = \frac{tan.A + tan.B}{1 - tan \:A.tan \: B } \\ \\ hence \: accordingly \\ = 1 - \frac{tan.45 + tan \: y}{1 - tan.45.tan.y} \\ \\ = 1 - \frac{1 + tan.y}{1 - 1 \times tan.y} \\ \\ since \: tan \: 45 = 1 \\ \\ = 1 - \frac{1 + tan.y}{1 - tan.y} \\ \\ = \frac{1 - tan.y - (1 + tan.y)}{1 - tan.y} \\ \\ = \frac{1 - tan.y - 1 - tan.y}{1 - tan.y} \\ \\ = \frac{ - 2.tan \: y}{1 - tan \: y} \\ \\ ie \: \: tan \: y \: (1 + tan \: y) = \frac{ - 2 \: tan.y}{1 - tan \: y} \\ \\ = 1 + tan.y = \frac{ - 2}{(1 - tan \: y)} \\ \\ (1 + tan \: y)(1 - tan \: y) = - 2 \\ 1 - tan {}^{2} y = - 2 \: \\ ie \: \: tan {}^{2} y = 3$

$taking \: square \: roots \: on \: both \: sides \\ we \: get \\ tan \: y = \sqrt{3} \\ so \: we \: know \: that \\ tan \: 60 \: \: ie \: \: tan \: \frac{\pi}{3} = \sqrt{3} \\ \\ ie \: y = \frac{\pi}{3} \\ \\ \: or \: \: y = 60$

$substitute \: value \: of \: y \: in \: (1) \\ we \: get \\ x = 105 \\ \\ ie \: \: x = -\frac{7.\pi}{12}$