Prove the Trig Identity: Grade twelve math

[Relz]

I used to be quite good at trig identities but it's been over a year and apparently i've lost that skill. I have quite a few questions but I will only post a couple for now:

1. (1/1+sin x) + (1/1-sinx) = 2 sec²x

2. tanx = (sin x + sin²x) / (cos x (1+ sin x))

 

[mmm4444bot]

(1/1+sin x) + (1/1-sinx) = 2 sec²x

Why do you type grouping symbols around the entire ratio? The purpose of those grouping symbols is to show everybody what the denominator is. In other words, everything inside the parentheses is the denominator. So, please text it that way.

1/(1 + sin[x]) + 1/(1 - sin[x]) = 2 sec²[x]

Add the ratios, on the left-hand side. On the right-hand side, rewrite the secant in terms of cosine. See if that helps spark any ideas.

 

[mmm4444bot]

I don't think that you even tried the second one. Factor the numerator, on the right-hand side. The next step should be obvious.

 

[Relz]

I don't think that you even tried the second one. Factor the numerator, on the right-hand side. The next step should be obvious.

Actually I have tried it but I get (sin x + sin x) /(cos x + cos x). This leaves me with 2 tanx, no? Am I allowed to just divide by two or did I completely screw this up? :$

 

[mmm4444bot]

Actually I have tried it but I get (sin x + sin x) /(cos x + cos x).

That statement above is what you should have said for exercise (2) in your original post. Now, I understand your error. You did not properly factor the numerator.

The numerator is sin[x] + sin[x]*sin[x]

There is a common factor on each side of the plus sign; the common factor is sin[x].

Factor out this common factor:

sin[x] * ( ? + ? )