Trig Properties / Algebra question

Audrey7

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May 6, 2013
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Hey everyone!

I'm in the middle of a homework that involves using trig properties and algebra to "prove" things, such as showing that sec(x) - cos(x) = sin(x)tan(x)

I didn't have any problems with questions like the one above, but I got stuck at the following:
tan((pi/2)-x) * tan(x) = 1

I know that sin(x) = (cos(x)-(pi/2)), but that's the only other time I've seen (pi/2) as part of the trig properties.

What am I supposed to do here? I got it to tan((pi/2)-x) = cot(x), but I don't understand why these are equal.

Any help is appreciated! Thanks.
 
You do know that pi/2 radians = 90° and that the sum of all three angles of a triangle is 180° = pi radians, don't you?
Trigonometric functions are defined as ratios between the sides in a right-angled triangle. Since one of the angles is pi/2 the sum of the other two must be pi/2. Can you take it from there?
 
Swapping from "\(\displaystyle \theta\)" to "\(\displaystyle \frac{\pi}{2}- \theta\)" is the same as switching from one acute angle in a right triangle to the other so is the same as swapping "near side" and "opposite side". Since each trig function is connected to its "co" function by swapping "near side" and "opposite side", what we get is just what you say: \(\displaystyle sin(\pi/2- \theta)= cos(\pi/2- \theta)\), \(\displaystyle tan(\theta)= cot(\pi/2- \theta)\), \(\displaystyle sec(\theta)= csc(\pi/2- \theta)\) and vice-versa.
 
...I got stuck at the following:
tan((pi/2)-x) * tan(x) = 1
Are you allowed to use the fact that \(\displaystyle \tan\left(\dfrac{\pi}{2}\, -\, x\right)\, =\, -\cot(x)\)? If so, then the result should follow fairly easily. ;)
 
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