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# How would I go about simplifying some of these trigonometric problems?

1. (tanx / cotx) + 1

2. (1-2cos^(x) / (2sinxcosx)

3. 1 - (sin^2(x)) / (1+cosx)

4. sin^2(x) * sec^2(X) * csc^2(x) * cos^2(x)

5.sinx * secx

6. (1-sin^2(x) * (1+tan^2(x)

7. (1-cosx) * (1+secx) * cotx

8. (csc^2(x)) * (1-cos^2(x))

9. (sinx / cscx) + (cosx / secx)

10. (1 / 1-sinx) / (1 / 1+sinx)

lastly, 11. tan^2(x) * csc^2(x) * cot^2(x) * sin^2(x)

I would appreciate if you included how you simplified the question.

Thanks,

Best answer will receive 10 points!

### 2 Answers

- spiffin456Lv 71 decade agoFavorite Answer
11. tan^2(x) * csc^2(x) * cot^2(x) * sin^2(x)

Because cot^2(x)=1/tan^2(x) and

csc^2(x)=1/sin^2(x), the answer is simply 1

- pink butterflyLv 51 decade ago
We could be here all day simplifying all 11. It sounds to me like you are too lazy to do it yourself.

All you have to remember is

tan=sin/cos

cot=cos/sin or 1/tan

sec=1/cos

csc=1/sin

cos^2x+sin^2x=1

1+tan^2x=sec^2x

1+cot^2x=csc^2x

Then if you have 2 fractions divided by each other, like you will in the first problem, then the second fraction gets flipped and you times it by the first.