Question

Need help with proving trig identity. Question is in Attached Photo

Answer 1

Here are the identities we can use

Answer 2

I got the first half of the identity
\[\sin(\frac{ \pi }{ 2 }-x)= cos x\]
but i dont know what to do with the other part

Answer 3

the cot((pi/2(+x) part

Answer 4

is cot((pi/2)+x)= 1/(tan((pi/2)+x)

Answer 5

\[\cot (\frac{ \pi }{ 2 }+x)=-\tan x\]

Answer 6

how would you prove that?

Answer 7

@mathmath333 @cwrw238 can u please help me

Answer 8

does
\[\cot(\frac{ \pi }{ 2 }+x)=\frac{ 1 }{ \tan(\frac{ \pi }{ 2 +x}) }\]

Answer 9

\(\large\tt \begin{align} \color{black}{sin(\dfrac{\pi}{2}-x)cot(\dfrac{\pi}{2}+x)\\
=cosx\times \dfrac{cos(\dfrac{\pi}{2}+x)}{sin(\dfrac{\pi}{2}+x)}\\
=cosx\times \dfrac{cos\dfrac{\pi}{2}cosx-sin\dfrac{\pi}{2} sinx}{sin\dfrac{\pi}{2}cosx+sinxcos\dfrac{\pi}{2}}\\
=cosx\times \dfrac{(0-sinx)}{cosx}\\
=-sinx\\
}\end{align}\)

Answer 10

Thank you

Answer 11

@mathmath333 you're supposed to teach. Not give solutions.

Answer 12

i did teach i didnt give only answers @calculusfunctions