View Full Version : How to establish these identities algebraically?
SoloDolo
Sep 29, 2010, 02:53 AM
1. Csc theta + cot theta = sin theta/ (1-cos theta)
2. csc x - sec x = (cos x-sin x)/(cosxsinx)
3. (cos x/(1+sinx))+((1+sinx)/cos x) = 2secx
Unknown008
Sep 29, 2010, 03:10 AM
For cosec, cot and sec, try expressing them in terms of cos and sin only. Try them now, and tell us what you get. These should be quite easy.
1. You see that two terms become one. Convert csc and cot into sin and cos, and mix the fractions.
2. Same thing.
3. Mix the fractions, and simplify as far as you can. The last term should give you sec.
Try it out and post what you get, where you are stuck :)
SoloDolo
Sep 29, 2010, 03:30 AM
Well for the first one I got as far as
1/sin + (cos/sin)= sin/(1-cos) then I'm not sure what to do.
The second one I got as far as
1/sin - 1/cos= (cos x-sin x)/(cosxsinx)
The third one I have no idea. The questions also instruct me to only work on one side of the identity. And write the input for each trig function. I'm not sure exactly what that means. Just started trig and I hate it! Haha
Unknown008
Sep 29, 2010, 03:44 AM
1.
\frac{1}{sin\theta} + \frac{cos\theta}{sin\theta} = \frac{1+cos\theta}{sin\theta}
Multiply by \frac{\sin\theta}{sin\theta}
Then, use: sin^2\theta = 1-cos^2\theta
Then, factorise the denominator using the fact that: a^2 - b^2 = (a+b)(a-b)
Try it now :)
2. It's the same as:
\frac{1}{a}- \frac{1}{b} = \frac{b}{ab} - \frac{a}{ab} = \frac{b-a}{ab}
3. Mix into a single fraction:
\frac{cos\theta}{1+sin\theta} + \frac{1+sin\theta}{cos\theta} = \frac{cos^2\theta + (1+sin\theta)^2}{cos\theta(1+sin\theta)}
Now, you simplify and continue. Post what you get!:)