37)
[tex]\frac{\sec x}{\tan x+\cot x}[/tex]First, we will simplify the denominator
tanx + cot x
tanx = sinx/cosx and cotx = cosx / sinx
[tex]\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{\sin ^2x+\cos ^2x}{\text{cosxsinx}}[/tex]sin²x +cos²x = 1
[tex]=\frac{1}{\cos x\sin x}[/tex]substitute back to the original expression
[tex]\frac{\sec x}{\cos x\sin x}[/tex]but secx = 1/cosx
[tex]\frac{1}{\cos x}dividedby\frac{1}{\cos x\sin x}[/tex][tex]=\frac{1}{\cos x}\times\cos x\sin x[/tex][tex]=\sin x[/tex]39)
[tex]\frac{1-\sin ^2x}{\cos x}[/tex]1 - sin²x = cos²x
substitute the above into the original expression
[tex]\frac{\cos ^2x}{\cos x}[/tex][tex]=\cos \text{ x}[/tex]40)
[tex]\frac{\csc ^2x-1}{\csc ^2x}[/tex]csc²x - 1 = cot²x
substitute the above into the original expression
[tex]\frac{\cot^2x}{\csc^2x}[/tex]But 1 + tan²x = cot²x
1 + tan²x
= 1 + sin²x/cos²x
= sin²x+cos²x /cos²x
=1 /cos²x
substitute into the expression
[tex]\frac{\frac{1}{\cos^2x}}{\csc ^2x}[/tex]1/cos²x ÷ csc²x
but csc²x = 1/sin²x
1/cos²x ÷ 1/sin²x
1/cos²x (sin²x)
[tex]\frac{\sin ^2x}{\cos ^2x}[/tex][tex]=\tan ^2x[/tex]38)
[tex]\frac{1+\sin x}{\cos x}+\frac{\cos x}{\sin x-1}[/tex]Find the lcm
[tex]\frac{(\sin x-1)(1+\sin x)+(\cos x)(\cos x)}{\cos x(\sin x-1)}[/tex][tex]\frac{\sin x+\sin ^2x-1-\sin x+\cos ^2x}{\cos x(\sin x-1)}[/tex]Re-arrange the numerator
[tex]\frac{\sin x-\sin x-1+\sin ^2x+\cos ^2x}{\cos x(\sin x-1)}[/tex][tex]\frac{-1+\sin ^2x+\cos ^2x}{\cos x(\sin x-1)}[/tex]But, sin²x+cos²x=1
[tex]\frac{-1+1}{\cos x(\sin x-1)}[/tex]= 0
