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if csc theta=-5/3 and theta has its terminal side in quadrant 3, find the exact value of tan 2 theta
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\[\tan 2x=\frac{ 2tanx }{ 1-\tan^2x }\]
\[sinx=\frac{ -3 }{ 5 }\]
now get the value of tan x and substitute in my formula
@mustafa2014 \[\csc x=\frac{ 1 }{ \sin x }\]
yep :)
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how do you find the value of tan x?
\[tanx=\frac{ \sin x }{ \cos x }\]
\[\cos x=\frac{ -4 }{ 5 }\]
so tan x= 3/4 right?
right
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so 279.5º?
the angle should be between 180 and 270
217 degrees approx.
I must of done my math wrong then. I plugged in 3/4. so 2tan(3/4)/1-tan^2(3/4) which was 1.8/0.36= 4.87* 180/π. what did i do wrong?
tan 2x=24/7
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just substitute the value of tanx=3/4 in the formula i gave in the beginning!
oh ok, I see. Thank you for your help again!
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