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# Trigonometric identities

(tanv+1)^2(tanv-1)^2=2sec^2v

### 2 Answers by Expert Tutors

John M. | Analytical assistance -- Writing, Math, and moreAnalytical assistance -- Writing, Math, ...
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Kayla,

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.
Alex S. | Retired, Looking to Share My Love of Mathematics and ComputersRetired, Looking to Share My Love of Mat...
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(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