The expression (cscx + cotx)^2 is the same as _____.
A. csc^2x + cot^2x
B. 1 + 2cot^2x + 2cscx cotx
C. 1 + 2csc^2x
D. csc^2x + 2cscx + cot^2x

Question

The expression (cscx + cotx)^2 is the same as _____.
  		A. 	csc^2x + cot^2x
  		B. 	1 + 2cot^2x + 2cscx cotx
  		C. 	1 + 2csc^2x
  		D. 	csc^2x + 2cscx + cot^2x

anonymous · Answer

U might as well just feed these things straight into Wolfram...

anonymous · Answer

http://www.wolframalpha.com/input/?i=%28cscx+%2B+cotx%29^2+

dumbcow · Answer

$$(\frac{1}{\sin x} + \frac{\cos x}{\sin x})^{2} = (\frac{1+\cos x}{\sin x})^{2} = \frac{\cos^{2} x + 2\cos x + 1}{\sin^{2} x} = \cot^{2} x +2\cot xcsc x +\csc^{2} x$$

dumbcow · Answer

its close to D but not exact, are the answer options correct

myininaya · Answer

cow you did it the long way lol

dumbcow · Answer

yeah im wierd like that

anonymous · Answer

so id go with d?

myininaya · Answer

don't go with any of them 
none of them are correct unless you mistyped something

anonymous · Answer

everythings right

myininaya · Answer

then write none of these and put what cow got

myininaya · Answer

wait

myininaya · Answer

$$1+\cot^2x=\csc^2x$$
$$\cot^2x+2cotxcscx+\csc^2x=\cot^2x+2cotxscscx+1+\cot^2x=2\cot^2x+2cotxcscx+1$$

anonymous · Answer

Problem is u can often write these things in a number of ways, takes some time to solve...

myininaya · Answer

i was wrong at first

dumbcow · Answer

wait its B)...yeah i just got that too

myininaya · Answer

i'm terrible

myininaya · Answer

we could had used that identity 1+cot^2x=csc^2x

dumbcow · Answer

trig can be tricky cause you can write the same thing different ways