Question #bd476

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Answer 1 · 2017-08-17T23:08:29

$tan (x) = \pm 1$ $tan(x)=+-1$

${sec (x)}^{2} = {tan (x)}^{2} + 1$ $sec(x)^2= tan(x)^2+1$

derived from ${sin (x)}^{2} + {cos (x)}^{2} = 1$ $sin(x)^2+cos(x)^2=1$ divide all by cos(x)^2

$2 {tan (x)}^{2} - ({tan (x)}^{2} + 1)$ $2tan(x)^2-(tan(x)^2+1)$

$2 {tan (x)}^{2} - {tan (x)}^{2} - 1 = 0$ $2tan(x)^2-tan(x)^2-1=0$

${tan (x)}^{2} = 1$ $tan(x)^2=1$

$tan (x) = \pm 1$ $tan(x)=+-1$

$tan (x)$ $tan(x)$ can be equal to

${- 45 \circ, 45 \circ, 135 \circ, 225 \circ} .$ ${ − 45 ∘ , 45 ∘ , 135 ∘ , 225 ∘ } .$