# Math Help - Trig identity problem

1. ## Trig identity problem

Prove -
1) $sec (u) + tan (u) = \frac {1} {sec(u) - tan(u)}$

2) $sec (u) + cosec (u)cot(u) = sec(u)cosec^2(u)$

3) $sin^2 (u)(1+sec^2 (u) ) = sec^2 (u) - cos^2 (u)$

4) $\frac {1-cos (u)} { sin(u)} = \frac {1} {cosec(u) + cot(u)}$

Any help appreciated

2. for number 2)
left hand side
= sec(u) + csc(u)cot(u)
= 1/cos(u) + [1/sin(u)]*[cos(u)/sin(u)]
= 1/cos(u) + [cos(u)/(sin(u))^2]
=
[(sin(u))^2 + (cos(u))^2]/[cos(u)*[sin(u)]^2]
= 1/
[cos(u)*[sin(u)]^2]
= sec(u)*[csc(u)]^2
= right hand side

3. Already solved these for you.

4. Originally Posted by jono
Prove -
1) $sec (u) + tan (u) = \frac {1} {sec(u) - tan(u)}$

2) $sec (u) + cosec (u)cot(u) = sec(u)cosec^2(u)$

3) $sin^2 (u)(1+sec^2 (u) ) = sec^2 (u) - cos^2 (u)$

4) $\frac {1-cos (u)} { sin(u)} = \frac {1} {cosec(u) + cot(u)}$

Any help appreciated
Okay, start with the Pythagorean theorem, which says $a^{2}+b^{2}=c^{2}$

Now, sin(x) is the opposite side over the hypotenuse. And because the trigonometric functions are defined by the unit circle, we can set our triangle on the unit circle. Since the radius of the unit circle is 1, our hypotenuse will be 1. So sin(x) = opposite side over 1, so it is just the opposite side. This means the sin(x) gives you the y value of your triangle (because the length of the opposite side is how tall the triangle is). And cos(x) = adjacent length over hypotenuse, and thus is just the length of your triangle.

So we see that we can plug these values into the Pythagorean theorem, length = cos(x), and height=sin(x)
$a^{2}+b^{2}=c^{2}\Rightarrow sin^{2}(x) + cos^{2}(x)=1$
(because 1 is the length of the radius on the unit circle, and that is our hypotenuse, and 1 squared is 1)

So our equation is
$sin^{2}(x) + cos^{2}(x)=1$

1. Convert from trig functions to length ratios
$\frac{opp^{2}}{hyp^{2}} + \frac{adj^{2}}{hyp^{2}}=1$

2. Multiply by $hyp^{2}$
$opp^{2} + adj^{2}=hyp^{2}$

3. Divide by $adj^{2}$
$\frac{opp^{2}}{adj^{2}} + \frac{adj^{2}}{adj^{2}}=\frac{hyp^{2}}{adj^{2}}$

4. Convert back to trig functions
$tan^{2}(x) + 1=sec^{2}(x)$

5. Subtract $tan^{2}$
$1=sec^{2}(x)-tan^{2}(x)$

6. Because $a^{2}-b^{2} = (a+b)(a-b)$
$1=[sec(x) + tan(x)][sec(x)-tan(x)]$

7. And divide
$\frac{1}{sec(x)-tan(x)}=sec(x) + tan(x)$