Proving trigonometric functions

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  • #1
chwala
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Mod note: Moved from a technical math forum, so this post is missing the homework template
i am trying to prove that ##1/sec∅-tan∅ ≡ sec ∅ + tan∅##
this is how i attempted it, i tried to show that the left hand side is equal to the right......
## 1/ 1/(cos∅-sin∅)/cos∅##
where i end up with
## (cos∅)/(1-sin ∅)## where this is ≡ ## (1+sin ∅)/(cos∅)##
taking ## cos^2∅/1-sin∅## divide by ##1+sin∅/cos∅## is equal to 1 (this is my proof), my concern is can we reduce
and show that ## cos∅/1-sin∅≡1+sin∅/cos∅## without necessarily dividing as i did it
 
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  • #2
Mentallic
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Can you please rewrite that, and this time use brackets appropriately. Either use \frac{a}{b} in latex to write a fraction with numerator 'a' and denominator 'b', or simply throw brackets in when multiple terms are either in the numerator or denominator.

While I know that you intended to write

[tex]\frac{1}{\sec\theta -\tan\theta}=\sec\theta+\tan\theta[/tex]

I first interpreted what you wrote as

[tex]\frac{1}{\sec\theta} -\tan\theta=\sec\theta+\tan\theta[/tex]

where this is ≡ ## (1+sin ∅)/(cos∅)##
Which is then clearly equal to [itex]\sec\theta+\tan\theta[/itex].

taking ## cos^2∅/1-sin∅## divide by ##1+sin∅/cos∅## is equal to 1 (this is my proof), my concern is can we reduce
and show that ## cos∅/1-sin∅≡1+sin∅/cos∅## without necessarily dividing as i did it
Just multiply the numerator and denominator by [itex]1+\sin\theta[/itex].

[tex]\frac{\cos\theta}{1-\sin\theta}\cdot \frac{1+\sin\theta}{1+\sin\theta}[/tex]

[tex]\frac{\cos\theta(1+\sin\theta)}{1-\sin^2\theta}[/tex]

[tex]\frac{\cos\theta(1+\sin\theta)}{\cos^2\theta}[/tex]
 
  • #3
chwala
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Thanks Mentallic i like Maths;)
 
  • #4
chwala
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thanks
 
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  • #5
Mentallic
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You're welcome :)

Also, notice something. We want to prove that

[tex]\frac{1}{\sec\theta-\tan\theta}=\sec\theta+\tan\theta[/tex]

So what if we simply multiplied the LHS by the conjugate of the denominator, resulting in

[tex]\frac{1}{\sec\theta-\tan\theta}\cdot \frac{\sec\theta+\tan\theta}{\sec\theta+\tan\theta}[/tex]

[tex]=\frac{\sec\theta+\tan\theta}{\sec^2\theta-\tan^2\theta}[/tex]

and at this point we know that it's supposed to be equal to the RHS (since you're expecting the result to be true), so clearly the denominator must be equal to 1. Well can we show that?

[tex]\sec^2\theta-\tan^2\theta=1[/tex] should remind you of the identity

[tex]\sec^2\theta=1+\tan^2\theta[/tex]

so if you mention this identity, then you can say that the denominator is equal to 1 and hence you've proved it without ever converting to sin and cos.
 

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