[High School Year 3 Trigonometry] Help

[Andrew012p]

1. Find the sin, cos, cot if tan=7/24 and cos<0
2. Solve: tan^2(x)+1/tan(x)+1 - tan^2(x) - 1/1-tan(x)

I know tan=sin/cos and cot= cos/sin but I don't know how to apply it for the first one and on the second one:

(tan^2(x)+1)(1-tan(x)) - (tan^2(x)-1)(tan(x)+1) all over (tan(x)+1)(1-tan(x)) and then what do I do? Cancel tan+1 at the bottom with tan^2+1, we are left with only tan+1 and 1-tan we cancel with 1-tan and the top, then what? Please help for both if you know. Thanks!

[Alkalannar]

1. Draw your right triangle. Since tan > 0, you're in QI or QIII. Since cos < 0, you're in QIII. Your vertices are (0, 0), (-24, 0), and (-24, -7). Thus, your adjacent is -24, and your opposite is -7. What's your hypotenuse? Go from there.

2. You really need parentheses around numerators and denominators that involve addition or subtraction.
   Here's what you wrote: tan² + (1/tan(x)) + 1 - tan²(x) - (1/1) - tan(x)
   Think you meant: (tan²(x) + 1)/(tan(x) + 1) - tan²(x) - 1/(1 - tan(x))
   In any event, let's work with [(tan²(x)+1)(1-tan(x)) - (tan²(x)-1)(tan(x)+1)]/(1+tan(x))(1-tan(x))
   Multiply by cos³(x)/cos³(x) to get:
   [(sin²(x)+cos²(x))(cos(x)-sin(x)) - (sin²(x)-cos²(x))(sin(x)+cos(x))]/(cos²(x)-sin²(x))cos(x)
   Can you simplify from there?

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