Question

How would you go about establishing the identity of (tan u)(cot u)-cos^2u=sin^2u.

Answer 1

do you know what ahemm say \(\bf sin^2(\theta)+cos^2(\theta)=?\)

Answer 2

Yes, the Pythagorean identity that equals one.

Answer 3

and what does tan*cot equal?

Answer 4

1?

Answer 5

well.... yes is 1 so \(\bf sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)={\color{brown}{ 1-cos^2(\theta)}}
\\ \quad \\
cot(\theta)={\color{purple}{ \cfrac{1}{tan(\theta)}}}
\\ \quad \\ \quad \\


tan(u)cot(u)-cos^2(u)=sin^2(u)\\ \quad \\ \cancel{ tan(u) }\cdot {\color{purple}{ \cfrac{1}{\cancel{ tan(u) }}}}-cos^2(u)=sin^2(u)
\\ \quad \\
{\color{brown}{ 1-cos^2(u)}}=sin^2(u)\)

Answer 6

Okay, that makes sense. I worked it out in a somewhat similar way, but I messed up on the step where you crossed out the tan u. Thank you so much for showing me how to find the other side.