Question

Prove this trig identity
2csc(x)tan(.5x)-1=tan^2(.5x)

Answer 1

Here's a better way to show it
\[2\csc x \tan \frac{ 1 }{ 2 }x=\tan ^{2}\frac{ 1 }{ 2 }x\]

Answer 2

Oh sorry should have a -1 on the LHS

Answer 3

hold the mayo

Answer 4

ahh rats

Answer 5

Thanks so much :)
that's all I need

Answer 6

\(\bf 2{\color{brown}{ csc(x) }} tan\left( \frac{x}{2} \right)-1=tan^2\left( \frac{x}{2}\right)
\\ \quad \\ \quad \\
2 {\color{brown}{\left[ csc\left( 2\cdot \frac{x}{2} \right) \right] }} tan\left( \frac{x}{2} \right)-1=tan^2\left( \frac{x}{2}\right)
\\ \quad \\
2 {\color{brown}{ \left[ \cfrac{1}{sin\left( 2\cdot \frac{x}{2} \right)} \right] }}tan\left( \frac{x}{2} \right)-1=tan^2\left( \frac{x}{2}\right)
\\ \quad \\
\textit{now, using double identies for sine, we get}
\\ \quad \\
2 {\color{brown}{ \left[ \cfrac{1}{
2sin\left(\frac{x}{2} \right)cos\left(\frac{x}{2}\right)} \right] }}tan\left( \frac{x}{2} \right)-1=tan^2\left( \frac{x}{2}\right)
\\ \quad \\
\cancel{2}\left[ \cfrac{1}{
\cancel{2} \cancel{sin\left(\frac{x}{2} \right)} cos\left(\frac{x}{2}\right)} \right] {\color{brown}{ \cfrac{\cancel{sin\left(\frac{x}{2} \right)}}{cos\left(\frac{x}{2}\right)} }}-1=tan^2\left( \frac{x}{2}\right)\)


there, see what you get
check the pythagorean identities :)

Answer 7

yw