Question

Prove that: \[\tan \theta \sin \theta + \cos \theta = \sec \theta\]

Answer 1

how should i start this?

Answer 2

use tan = y/x
sin= y/r
cos= x/r
Plug into your equation and such

Answer 3

http://www.thestudentroom.co.uk/showthread.php?t=130710 it will help you :)

Answer 4

\[(\frac{y}{x})(\frac{y}{r}) + \frac{x}{r}?\]

Answer 5

change tanx to sinx/cosx and then multiply by sinx, then find a common denominator to add cosx.

Answer 6

sec is is r/x oops lol

Well my reasoning is that i hate tri and i work better with variables like x,y and r instead of sines cosines and tans

Answer 7

hmm i think those doesnt make sense guys @_@

Answer 8

\[\frac{\sin(\theta)}{\cos(\theta)} \cdot \sin(\theta)+\cos(\theta)=\frac{\sin^2(\theta)}{\cos(\theta)}+\frac{\cos(\theta)}{\cos(\theta)} \cdot \cos(\theta)\]
I multiply second term by cos(theta)/cos(theta)
so now you can combine the fractions
and guess what you get on top? :)

Answer 9

and then sin^2 + cos^2 = 1 and 1/cosx = secx.

Answer 10

i dont get what you did @myininaya o.O

Answer 11

\[\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}\]

Answer 12

First step I did was replace tan(theta) with that

Answer 13

Second step multiply second term by 1=cos(theta)/cos(theta)

Answer 14

but why?

Answer 15

Because you have two terms on one side and only one on the other

Answer 16

If you combine the fractions on the side with two then you will have one term

Answer 17

oh i see...for cmmon denominator?

Answer 18

yes

Answer 19

so then i'll have \[\frac{sin^2 \theta + \cos^2 \theta}{\cos \theta} = \frac{1}{\cos \theta} = \sec \theta\] COOL! *_*

Answer 20

yep