Trigonometric identities

Question

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

Answer 1

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). 

Answer 2

(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