show that(tanx)/(1-cotx) + (cotx)/(1-tanx) = 1 + secxcscx

Question

pre calculus

Answer 1

Here is one possible way to prove this trig. identity:

[tanx / (1-cotx)] + [cotx / (1-tanx)] *Given

[tanx / (1 - (1/tanx))] + [(1/tanx) / (1-tanx)] *Substitute cotx = 1/tanx

[tan²x / (tanx -1)] + [1 / (tanx (1-tanx))] *Simplify the fractions in each term

[tan²x / (tanx -1)] - [1 / (tanx(tanx-1))] * 1-tanx = -(tanx -1), distributive property

[tan³x - 1] / [tanx (tanx-1)] * Fraction subtraction

[(tanx - 1)(tan²x + tanx + 1)] / [tanx(tanx-1)] * Difference of cubes

tanx + 1 + cotx * Simplify

1 + [sinx/cosx] + [cosx/sinx] * Substitute tan = sin/cos and cot = cos/sin

1 + [(sin²x + cos²x) / (sinxcosx)] * Fraction addition

1 + [1/(sinxcosx)] *Substitute sin² + cos² = 1

1 + secxcsx * Substitute sec = 1/cos and csc = 1/sin. DONE :)

Hope this helps! Of course, there are multiple ways to prove trig. identities. It'd be awesome if someone could come up with a shorter proof to post :)