Question

Simplify:
Sin(180˚-θ)/sin(90˚-θ) x tan(90˚-θ)

Answer 1

is this the correct equation of your question?
\[\frac{ \sin(180-\theta) }{ \sin(90-\theta)(\tan(90-\theta)) }\]

Answer 2

if it is, then we can start proving this.

Answer 3

no sorry

Answer 4

is there an x in the denominator?

Answer 5

Lemme know so I fix the equation.

Answer 6

oh I see
\[\frac{ \sin(180-\theta) }{ \sin(90-\theta) } (\tan(90-\theta))\]

Answer 7

Sin(180-theta) ? sin(90-theta) multiplied by tan (90-theta)

Answer 8

like that right?

Answer 9

yes

Answer 10

Ok, do you know the cofunction identities?

\[\sin(90-\theta)= \cos ( \theta )\]

Answer 11

90 degrees = pi/2
same as
\[\sin(\frac{ \pi }{ 2 } - \theta) = \cos(\theta)\]

Answer 12

and
\[\sin(180-\theta) = \sin (\theta)\]

Answer 13

yes

Answer 14

and
\[\tan(90-\theta)= \cot (\theta)\]

Answer 15

so is that the final answer?

Answer 16

remember that sin/cos= tan

Answer 17

oh ok....Thanks heaps

Answer 18

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Answer 19

That helps a lot with other questions as well thank you

Answer 20

hm wait, I think I have a small error

Answer 21

when I plug in it was supposed to be
\[\frac{ \sin(x) }{ \cos(x) } (\cot(x))\]
\[=\tan(x) (\frac{ 1 }{ \tan(x) })\]
= 1.

tan(90-theta) = cot (theta)

Answer 22

Yes that makes sense........thanks

Answer 23

:)

Answer 24

nice work...

Answer 25

like how I tagged pantie power instead of you paki, @paki
wow..

Answer 26

:)

Answer 27

hahahha yeah :)