Question

(160(1-sin x)²)(1 + (tan x)²)

Answer 1

@aml13 Are you supposed to solve for 'x' or just simplify the expression?

Answer 2

simplify!

Answer 3

Notice that \(\bf 1+tan^2(x)=sec^2(x)=\frac{1}{cos^2(x)}\). Hence:\[\frac{160-320\sin(x)+160\sin^2(x)}{\cos^2(x)}=160\sec^2(x)-320\tan(x)\sec(x)+160\tan^2(x)\]Factoring by grouping yields:\[\bf =160\sec^2(x)-160\sec(x)\tan(x)-160\sec(x)\tan(x)+160\tan^2(x)\]\[\bf = 160\sec(x)(\sec(x)-\tan(x))-160\tan(x)(\sec(x)-\tan(x))\]\[\bf = (160\sec(x)-160\tan(x))(\sec(x)-\tan(x))\]\[\bf = 160[\sec(x)-\tan(x)]^2\]
You can expand the final form if desired. @aml13

Answer 4

thank you!!