Question

Verify the identity. Show your work.
(1 + tan2u)(1 - sin2u) = 1

Answer 1

\(\large\color{black}{ \displaystyle(1+\tan~2u)(1-\sin~2u)=1 }\)

Expand the parenthesis, I suppose.

Answer 2

I don't think that it is indeed an identity (\(\forall x\)) to be honest, but ...

Answer 3

@q12157

Answer 4

Since sin(2u)=2sinucosu and cos2u=1-2sin^2(u),
cos2u is also equal to cos^2(u)-sin^2(u) or 2cos^2(u)-1

Answer 5

you use those identities and plug them in...expand and then voila you should get 1 on the left side.

Answer 6

Wait, is it?

\(\large\color{black}{ \displaystyle(1+\tan^2u)(1-\sin^2u)=1 }\)

I am asking, because THIS would be an indetity as opposed to the first one.

Answer 7

\(\large\color{black}{ \displaystyle(1+\tan^2u)(1-\sin^2u)=1 }\)

Some hepful properties:
\(\large\color{blue}{ \displaystyle[1]\quad \quad \sec^2\theta =1+\tan^2\theta }\)
\(\large\color{blue}{ \displaystyle[2]\quad \quad \cos^2\theta=1-\sin^2\theta }\)
\(\large\color{blue}{ \displaystyle[3]\quad \quad 1= \cos\theta \times\sec \theta \quad \Longrightarrow 1=\cos^{\color{red}N}\theta \times\sec^{\color{red}N}\theta}\)