Question

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

Answer 1

 

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

$\displaystyle\\=-1- \frac{\sin^2x}{\cos^2x}+\frac{\cos^2x}{\sin^2x}+ \frac{1}{\cos^2x}- \frac{1}{\sin^2x} +2=\\\\=-\frac{\sin^2x }{\cos^2x}+\frac{\cos^2x}{\sin^2x}+ \frac{1}{\cos^2x}- \frac{1}{\sin^2x} +2-1=\\\\=\frac{-\sin^4x+\cos^4x}{sin^2x\cos^2x}+\frac{\sin^2x-\cos^2x}{\sin^2x\cos^2x}+1=\\\\=\frac{\cos^4x-\sin^4x}{\sin^2x\cos^2x}-\frac{\cos^2x-\sin^2x}{\sin^2x\cos^2x}+1=\\\\=\frac{(\cos^2x+\sin^2x)(\cos^2x-\sin^2x)}{\sin^2x\cos^2x}-\frac{\cos^2x-\sin^2x}{\sin^2x\cos^2x}+1=$

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