Pre-Calc

posted by .

Use the fundamental identities and appropriate algebraic operations to simplify the expression.

cosx / (1+sinx)

Answer: secx

Can someone please assist as to how I get to the answer?

  • Pre-Calc -

    You must have a typo

    cosx/(1+sinx) is NOT equal to secx

    try it with any angle.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Pre-Calc

    Trigonometric Identities Prove: (tanx + secx -1)/(tanx - secx + 1)= tanx + secx My work so far: (sinx/cosx + 1/cosx + cosx/cosx)/(sinx/cos x - 1/cosx + cosx/cosx)= tanx + cosx (just working on the left side) ((sinx + 1 - cosx)/cosx)/((sinx …
  2. Simplifying with Trigonometry Identities

    Hi, I am a senior in High School having a really difficult time with two problems. I have to prove using the trigonometric identities that they equal each other but I am having a really hard time trying to get them to equal each other. …
  3. Pre-calc

    prove the identity: (cosx)(tanx + sinx cotx)=sinx+cos(squared)x i need steps to show how i got the answer generally, it is a good idea to change all trig ratios to sines and cosines, and start with the more complicated-looking side. …
  4. Math

    Use the fundamental identities to simplify the expression. csc Q / sec Q so this would be 1/sinx / 1/cosx and then 1/sinx times cosx/1 = sin x cosx Is this correct so far?
  5. Math(Please help)

    Use the fundamental identities to simplify the expression. csc Q / sec Q so this would be 1/sinx / 1/cosx and then 1/sinx times cosx/1 = sin x cosx Is this correct so far?
  6. Pre-Cal

    Use the fundamental identities to simplify the expression. csc Q / sec Q so this would be 1/sinx / 1/cosx and then 1/sinx times cosx/1 = sin x cosx Is this correct so far?
  7. Trigonometry

    Prove the following trigonometric identities. please give a detailed answer because I don't understand this at all. a. sin(x)tan(x)=cos(x)/cot^2 (x) b. (1+tanx)^2=sec^2 (x)+2tan(x) c. 1/sin(x) + 1/cos(x) = (cosx+sinx)(secx)(cscx) d. …
  8. Precalculus/Trig

    I can't seem to prove these trig identities and would really appreciate help: 1. cosx + 1/sin^3x = cscx/1 - cosx I changed the 1: cosx/sin^3x + sin^3x/sin^3x = cscx/1-cosx Simplified: cosx + sin^3x/sin^3x = cscx/1-cosx I don't know …
  9. Calculus 2 Trigonometric Substitution

    I'm working this problem: ∫ [1-tan^2 (x)] / [sec^2 (x)] dx ∫(1/secx)-[(sin^2x/cos^2x)/(1/cosx) ∫cosx-sinx(sinx/cosx) ∫cosx-∫sin^2(x)/cosx sinx-∫(1-cos^2(x))/cosx sinx-∫(1/cosx)-cosx sinx-∫secx-∫cosx …
  10. Gr 11 Functions

    Simplify (1+tanx)^2 The answer is (1-sinx)(1+sinx) Here's what I do: 1 + 2tanx + tan^2x When I simplify it becomes 1 + 2(sinx/cosx) + (1+secx)(1-secx) What am I doing wrong?

More Similar Questions