Question

Simplify -sin^2x-cos^2x-tan^2x+cot^2x+sec^2x-csc^2x+2

Answer 1

Answer: actually it is very complicated to explain but the answer is 1

Answer 2

`\displaystyle\\\text{simplify: } -\sin^2x-\cos^2x-\tan^2x+\cot^2x+\sec^2x-\csc^2x+2\\\\\text{We use the formulas:}\\1)~~\sin^2x+\cos^2x=1\\\\2)~~\tan x=(\sin x)/(\cos x) ~~~~~~~3)~~\cot x= (\cos x)/(\sin x)\\ \\4)~~\sec x=(1)/(\cos x) ~~~~~~~~5)~~\csc x=(1)/(\sin x)\\ \\\text{Answer:}\\\\-\sin^2x-\cos^2x-\tan^2x+\cot^2x+\sec^2x-\csc^2x+2=\\\\=-(\sin^2x+\cos^2x)- (\sin^2x)/(\cos^2x)+(\cos^2x)/(\sin^2x)+\sec^2x-\csc^2x+2=`

`\displaystyle\\=-1- (\sin^2x)/(\cos^2x)+(\cos^2x)/(\sin^2x)+ (1)/(\cos^2x)- (1)/(\sin^2x) +2=\\\\=-(\sin^2x )/(\cos^2x)+(\cos^2x)/(\sin^2x)+ (1)/(\cos^2x)- (1)/(\sin^2x) +2-1=\\\\=(-\sin^4x+\cos^4x)/(sin^2x\cos^2x)+(\sin^2x-\cos^2x)/(\sin^2x\cos^2x)+1=\\\\=(\cos^4x-\sin^4x)/(\sin^2x\cos^2x)-(\cos^2x-\sin^2x)/(\sin^2x\cos^2x)+1=\\\\=((\cos^2x+\sin^2x)(\cos^2x-\sin^2x))/(\sin^2x\cos^2x)-(\cos^2x-\sin^2x)/(\sin^2x\cos^2x)+1=`

`\displaystyle\\=((1)\cdot(\cos^2x-\sin^2x))/(\sin^2x\cos^2x)-(\cos^2x-\sin^2x)/(\sin^2x\cos^2x)+1=\\\\=\underbrace{(\cos^2x-\sin^2x)/(\sin^2x\cos^2x)-(\cos^2x-\sin^2x)/(\sin^2x\cos^2x)}_(=~0)\,+\,1=0+1=\boxed{\boxed{\bf1}}`