Given $cos x = - \frac{3}{5}$ $cosx = -3/5$ and $0 < x < 180$ $0<x<180$ , what is $tan (x + 45)$ $tan(x+45)$ ?

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Given $cos x = - \frac{3}{5}$ $cosx = -3/5$ and $0 < x < 180$ $0<x<180$ , what is $tan (x + 45)$ $tan(x+45)$ ?

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Answer 1 · 2018-04-16T10:43:39

${tan (x + 45)}^{\circ} = 1$ $tan(x+45)^@=1$

Explanation:

$using the trigonometric identities$ $"using the "color(blue)"trigonometric identities"$

$∙ x tan x = \frac{sin x}{cos x}$ $•color(white)(x)tanx=sinx/cosx$

$∙ x tan (x + y) = \frac{tan x + tan y}{1 - tan x tan y}$ $•color(white)(x)tan(x+y)=(tanx+tany)/(1-tanxtany)$

$∙ x {sin}^{2} x + {cos}^{2} x = 1$ $•color(white)(x)sin^2x+cos^2x=1$

$\Rightarrow sin x = \pm \sqrt{1 - {cos}^{2} x}$ $rArrsinx=+-sqrt(1-cos^2x)$

$cosx<0 and 0 < x < 180$ $"cosx<0" and "0 < x < 180$

$\Rightarrow x is in the second quadrant where sin x > 0$ $rArr" x is in the second quadrant where "sinx>0$

$sin x = + \sqrt{1 - {(- \frac{3}{5})}^{2}}$ $sinx=+sqrt(1-(-3/5)^2)$

$sin x = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$ $color(white)(sinx)=sqrt(1-9/25)=sqrt(16/25)=4/5$

$\Rightarrow tan x = \frac{4}{5} \times - \frac{5}{3} = - \frac{4}{3}$ $rArrtanx=4/5xx-5/3=-4/3$

$\Rightarrow tan (x + 45) = \frac{- \frac{4}{3} + 1}{1 - \frac{4}{3}} = \frac{- \frac{1}{3}}{- \frac{1}{3}} = 1$ $rArrtan(x+45)=(-4/3+1)/(1-4/3)=(-1/3)/(-1/3)=1$

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Given $cos x = - \frac{3}{5}$ $cosx = -3/5$ and $0 < x < 180$ $0<x<180$ , what is $tan (x + 45)$ $tan(x+45)$ ?

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Explanation:

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Given $cos x = - \frac{3}{5}$ $cosx = -3/5$ and $0 < x < 180$ $0<x<180$ , what is $tan (x + 45)$ $tan(x+45)$ ?