Question

Please do no 7 for me. I’m practising for my exam and can’t find the solution. The correct answer will get 15 points. No spam

Answer 1

Step-by-step explanation:

Prove that

$\frac{ \tan(x) }{ \sec(x) - 1 } + \frac{ \sin(x) }{1 + \cos(x) } = 2 \csc(x)$

Recall that ta x = sin x/ cos x and sec x = 1/cos x. So the 1st term on the LHS becomes

$\frac{ \frac{ \sin(x) }{ \cos(x) } }{ \frac{1}{ \cos(x) } - 1 } + \frac{ \sin(x) }{1 + \cos(x) } = 2 \csc(x)$

or with the cos x cancelling out, the equation becomes

$\frac{ \sin(x) }{1 - \cos(x) } + \frac{ \sin(x) }{1 + \cos(x) } = 2 \csc(x)$

Combining the terms on the LHS,

$\frac{2 \sin(x) }{(1 - \cos(x))(1 + \cos(x) ) } = 2 \csc(x)$

$\frac{2 \sin(x) }{1 - {( \cos(x) )}^{2} } = \frac{2 \sin(x) }{ { (\sin(x)) }^{2} } \\ or \\ \frac{2}{ \sin(x) } = 2 \csc(x)$