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

Answers

Answer 1

Answer:

$\displaystyle\n\text{simplify: } -\sin^2x-\cos^2x-\tan^2x+\cot^2x+\sec^2x-\csc^2x+2\n\n\text{We use the formulas:}\n1)~~\sin^2x+\cos^2x=1\n\n2)~~\tan x=(\sin x)/(\cos x) ~~~~~~~3)~~\cot x= (\cos x)/(\sin x)\n \n4)~~\sec x=(1)/(\cos x) ~~~~~~~~5)~~\csc x=(1)/(\sin x)\n \n\text{Answer:}\n\n-\sin^2x-\cos^2x-\tan^2x+\cot^2x+\sec^2x-\csc^2x+2=\n\n=-(\sin^2x+\cos^2x)- (\sin^2x)/(\cos^2x)+(\cos^2x)/(\sin^2x)+\sec^2x-\csc^2x+2=$

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

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

Answer 2

Answer:

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

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Answers

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Answers

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Answers

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Graph the system of inequalities presented here on your own paper, then use your graph to answer the following questions:y < 4x − 8
y is greater than or equal to negative 5 over 2 times x plus 5

Part A: Describe the graph of the system, including shading and the types of lines graphed. Provide a description of the solution area. (6 points)

Part B: Is the point (5, −8) included in the solution area for the system? Justify your answer mathematically. (4 points)