How do we know that Tan(180-x)=-tanx And Cos(180-x)=-cosx Sin(180-x)=Sinx Tan(90-x)=1/-tanx etc...is there any way we can find out these?

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How do we know that Tan(180-x)=-tanx And Cos(180-x)=-cosx Sin(180-x)=Sinx Tan(90-x)=1/-tanx etc...is there any way we can find out these?

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Answer 1 · 2018-04-20T23:54:00

See below

Explanation:

Let's try proving two them

$tan (180^{\circ} - x) = - tan x$ $tan(180^@-x)=-tanx$

Apply tangent sum identity:
$\frac{{tan 180}^{\circ} - tan x}{1 + {tan 180}^{\circ} tan x}$ $(tan180^@-tanx)/(1+tan180^@tanx)$

Simplify:
$\frac{0 - tan x}{1 + 0 \cdot tan x}$ $(0-tanx)/(1+0*tanx)$

$- \frac{tan x}{1} = - tan x$ $-tanx/1= -tanx$

The second one we will try is
$tan (90^{\circ} + x) = \frac{1}{- tan x}$ $tan(90^@+x)=1/-tanx$ :

Unfortunately $tan (90^{\circ})$ $tan(90^@)$ is undefined so:
Apply quotient identity:
$\frac{sin (90^{\circ} + x)}{cos (90^{\circ} + x)} = \frac{1}{- tan x}$ $sin(90^@+x)/cos(90^@+x)=1/-tanx$ :

Now apply the sine and cosine sum identities:
$\frac{{sin 90}^{\circ} cos x + {cos 90}^{\circ} sin x}{{cos 90}^{\circ} cos x - {sin 90}^{\circ} sin x} = \frac{1}{- tan x}$ $(sin90^@cosx+cos90^@sinx)/(cos90^@cosx-sin90^@sinx)=1/-tanx$

Simplify:
$\frac{(1) cos x + (0) sin x}{(0) cos x - (1) sin x} = \frac{1}{- tan x}$ $((1)cosx+(0)sinx)/((0)cosx-(1)sinx)=1/-tanx$

$\frac{cos x}{- sin x} = - \frac{1}{tan x}$ $cosx/-sinx=-1/tanx$

Apply quotient identity:
$- cot x = - \frac{1}{tan x}$ $-cotx=-1/tanx$

Apply reciprocal identity:
$- \frac{1}{tan x} = - \frac{1}{tan x}$ $-1/tanx=-1/tanx$

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How do we know that Tan(180-x)=-tanx And Cos(180-x)=-cosx Sin(180-x)=Sinx Tan(90-x)=1/-tanx etc...is there any way we can find out these?

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Explanation:

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