Question

Please help yall! this is driving me insane. pictures attached!

Answer 1

`(\cos x)/(1- \sin x)=\sec x + \tan x`

`(\cos x)/(1- \sin x)=\boxed{(1)/(\cos x) + (\sin x)/(\cos x)}`

`(\cos x)/(1- \sin x)=\boxed{(1+\sin x)/(\cos x)}`

`(\cos x)/(1- \sin x)=\boxed{(\cos x(1+\sin x))/(\cos^2 x)}`

`(\cos x)/(1- \sin x)=\boxed{(\cos x(1+\sin x))/(1-\sin^2 x)}`

`(\cos x)/(1- \sin x)=\boxed{(\cos x(1+\sin x))/((1-\sin x)(1+ \sin x))}`

`(\cos x)/(1- \sin x)=\boxed{(\cos x)/(1- \sin x)}`

Explanation:

Given equation:

`(\cos x)/(1- \sin x)=\sec x + \tan x`

`\textsf{Use\;the\;identities\;\;$\boxed{\sec x=(1)/(\cos x)}$\;\;and\;\;$\boxed{\tan x=(\sin x)/(\cos x)}$\;:}`

`\implies (\cos x)/(1- \sin x)=(1)/(\cos x) + (\sin x)/(\cos x)`

`\textsf{Apply the fraction rule:} \quad (a)/(c)+(b)/(c)=(a+b)/(c)`

`\implies (\cos x)/(1- \sin x)=(1+\sin x)/(\cos x)`

Multiply the numerator and denominator by cos x:

`\implies (\cos x)/(1- \sin x)=(\cos x(1+\sin x))/(\cos^2 x)`

`\textsf{Use\;the\;identity\;\;$\boxed{\sin^2x+\cos^2x=1}$\;\;to\;rewrite\;the\;denominator:}`

`\implies (\cos x)/(1- \sin x)=(\cos x(1+\sin x))/(1-\sin^2 x)`

`\textsf{Use\;the\;Difference\;of\;Two\;Squares\;formula\;\;$\boxed{a^2- b^2= (a - b)(a + b)}$}\\\textsf{to\;rewrite\;the\;denominator:}`

`\implies (\cos x)/(1- \sin x)=(\cos x(1+\sin x))/((1-\sin x)(1+ \sin x))`

Cancel the common term (1 + sin x):

`\implies (\cos x)/(1- \sin x)=(\cos x)/(1- \sin x)`

Thus proving the identity.