Trigonometry Identity Questions

Question

Hi, I have two problems that I cannot figure out how to even go about solving. I was hoping someone could help me with them.

1.) 9sec² θ - 5tan²θ = 5 + 4sec² θ

2.) (1 + sinx) / (1 - sin x) = (csc x + 1) / (csc x - 1)

I greatly appreciate the help. Thanks!

Answer 1

Problem 1) 9sec²θ - 5tan²θ = 5 + 4sec²θ start by subtracting 4sec²θ from both sides:

5sec²θ - 5tan²θ = 5 now factor out 5 on the left

5(sec²θ - tan²θ) = 5

Using 1 + tan²θ = sec²θ replace "sec²θ" with "1 + tan²θ" to get:

5(1 + tan²θ - tan²θ) = 5 or

5(1) = 5

5 = 5

Problem 2) (1 + sin(x)) / (1 - sin (x)) = (csc(x) + 1) / (csc(x) - 1)

Using csc(x) = 1/sin(x) replace the csc(x) on the right side to get:

(1 + sin(x)) / (1 - sin (x)) = (1/sin(x) + 1) / (1/sin(x) - 1)

get a common denominator on the right:

(1 + sin(x)) / (1 - sin (x)) = (1/sin(x) + sin(x)/sin(x)) / (1/sin(x) - sin(x)/sin(x))

(1 + sin(x)) / (1 - sin (x)) = (1 + sin(x))/sin(x)) / (1 - sin(x))/sin(x)

(1 + sin(x)) / (1 - sin (x)) = (1 + sin(x)) / (1 - sin(x))

Answer 2

For the first one, split the 9/cos² into 5/cos² + 4/cos² and it will come out easily.

In the second one, write the csc as 1/sin and it will come out almost directly!