[izunia27]

how do I solve these 2 proofs? I'm really confused :(... help!

cscϑ+cotϑ/tanϑ+sinϑ=cotϑcscϑ

(1+sinϑ/1-sinϑ)-(1-sinϑ/1+sinϑ)=4tanϑsecϑ

[Unknown008]

1.



You can start by expressing all those into the basic trig ratios, namely cos and sin.

2.



Start by combining the fractions.

[jcaron2]

For the first one, you first need to rewrite everything in terms of sines and cosines (that's almost always a good place to start with these types of problems). That will give you some fractional terms in both the numerator and denominator of the left side of the equation. Next you need to combine them into single fractions. For example, the numerator is csc + cot. Rewrite as 1/sin+cos/sin, then combine the two fractions to get (1+cos)/sin. Now you need to do the same for the denominator. That one's a little bit tougher because the two terms don't have a common denominator. You'll need to multiply the sin term by cos/cos so you can get a common denominator to be able to add the fractions. Once you do that, you should find that you can factor out a (1+cos). Finally, do the fractional division (by inverting and multiplying by the denominator). Then you should find that the (1+cos) terms cancel out, leaving you with a cos/sin^2. That, of course, is equal to the right side of the equation.

For the second one, you need to subtract the two fractions by finding a common denominator. That's easy enough to do by multiplying the first term by (1+sin)/(1+sin) and multiplying the second term by (1-sin)/(1-sin). That gives them both a denominator of (1+sin)(1-sin), which multiplies out to give you (1-sin^2). Do you recognize what this new denominator is equivalent to with a trig identity? Once you know that and you go through the tedious multiplications in the numerator, you should find that several terms in the numerator cancel and the right side of the equation will pop out.

[jcaron2]

LOL! You beat me to it.

[Unknown008]

Oops, I didn't know you were posting too :o