Ask a question
0 0

Trigonometric identities

Tutors, please sign in to answer this question.

2 Answers

The Pythagorean identity might be useful here rather than a double angle identity.
Remember that sin^2 + cos^2 = 1 and so if we divide both sides by cos^2, we get
sin^2/cos^2 + cos^2/cos^2 = 1/cos^2 or
tan^2 + 1 = sec^2
The (tan(v) + 1)^2 ≠ sec^2(v).  Rather it is equal to (tan(v) + 1)(tan(v) + 1), which you need to FOIL out; the same goes for the (tan(v) - 1)(tan(v) - 1). 


Check out my answer ... Note mistake (typo) in the original statement of the problem:
"I believe you left out a + (plus, rather than times/multiply) ... you meant
(tan v + 1) 2+ (tan v - 1)2= 2sec2 v"
I am unclear on hoe you propose to use Pythogorean theorem.
(tan v + 1)2 * (tan v - 1)2 = (tan2 v - 1)2
This NOT equal to 2sec2 v.
I believe you left out a + (plus, rather than times/multiply) ... you meant
(tan v + 1)2 + (tan v - 1)2 = 2sec2 v
This equality is correct:
(tan v +1)2 = tan2 v + 2(tan v) +1
(tan v - 1)2 = tan2 v - 2(tan v) +1
(tan v + 1)2+(tan v - 1)2=2tan2v + 2=2(tan2v + 1)
since tan2v+1 = sec2v
(tan v + 1)2 + (tan v - 1)2 = 2sec2v


I'm not sure on where to go with that. I went in a completely different direction when I tried to solve it. I kept the secant and turned. (tanv+1)^2 into (tan^2+1) and that would be sec^2. After that I was lost on what to do with (tan^2v-1) to get it to equal 2sec^2v. 
Why did you mention cos(2v)? The problem doesn't have double angles in it.
I misread the problem ... sec^2v as sec(2v), rather than sec2v.  I have revised my answer, based on correct reading ... and,  now provide correct solution.