1. ## trig identities

I have no idea how to prove these. Any solutions would be greatly appreciated.

1)
tan(x+y+z) = (tanx + tany +tanz - tanxtanytanz)/(1- tanxtany- tanxtanz-tanytanz)

2)
sin(pi/2 +x)cos(pi-x)cot(3pi/2+x)=sin(pi/2-x)sin(3pi/2-x)cot(pi/2+x)

3) csc^2(pi/2-x)=1+sin^2xcsc^2(pi/2-x)

2. Hello, anna12345!

$1)\;\;\tan(x+y+z) \:=\: \frac{\tan x + \tan y +\tan z - \tan x\tan y\tan z}{1- \tan x\tan y - \tan x\tan z - \tan y\tan z}$

We're expected to know: . $\tan(A + B) \;=\;\frac{\tan A + \tan B}{1 - \tan A\tan B}$
. . and be able to apply it twice.

$\tan\bigg[(x + y) + x\bigg] \;=\;\frac{\tan(x + y) + \tan z}{1 - \tan(x+y)\tan z} \;=\;\frac{\frac{\tan x+\tan y}{1-\tan x\tan y} + \tan x}{1 - \frac{\tan x +\tan y}{1 - \tan x\tan y}\tan z}$

Multiply top and bottom by $(1 - \tan x\tan y)\!:$

. . $\frac{(1 - \tan x\tan y)\left(\frac{\tan x+\tan y}{1-\tan x\tan y} + \tan x\right)} {(1-\tan x\tan y)\left(1 - \frac{\tan x +\tan y}{1 - \tan x\tan y}\tan z\right)}$ . $= \;\frac{\tan x + \tan y + \tan z(1 - \tan x\tan y)}{(1-\tan x\tan y) - (\tan x + \tan y)\tan x}$

. . $= \;\frac{\tan x + \tan y + \tan z - \tan x\tan y\tan z}{1 - \tan x\tan y - \tan x\tan z - \tan y\tan z}$

3. Thank you so much! It all makes sense to me now.

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# A B C=Ï€ tanx tany tanz= tanx.tany.tanz

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