3

Factor the expression and use the fundamental identities to simplify. There is more than one correct form of the answer. $$7 \sin^2 x \csc^2 x − 7 \sin^2 x$$

I'm reviewing for a test and going over my old homework, is 7 a possible solution (I'm asking because the last time I did this on my homework I got $7\cos(x)$?

$7 \sin^2 x (csc^2 x − 1)$

$=7 \sin^2 x (1 + \cot^2x)$

$=7 \sin^2 x (1 + \frac {\cos^2x}{\sin^2x})$

$=7 \sin^2 x (\frac{\sin^2x}{\sin^2x} + \frac {\cos^2x}{\sin^2x})$

$=7\sin^2x (\frac{\sin^2x+\cos^2x}{\sin^2x})$

$=7\sin^2x (\frac{1}{\sin^2x})$

$=\frac{7\sin^2x}{\sin^2x}$

$=7$

Harish Chandra Rajpoot
  • 37,450
TheNewGuy
  • 533

2 Answers2

2

$$7\sin^2 x\csc^2 x-7\sin^2 x=7\sin^2 x\frac{1}{\sin^2x}-7\sin^2 x=$$ $$=7-7\sin^2 x=7(1-\sin^2 x)=7\cos^2x$$

Adi Dani
  • 16,949
  • This is what should come to mind first. Why does anyone need to factor out the $7\sin^2x$ first? – najayaz Jan 17 '16 at 10:19
0

Notice, your mistake $\csc^2-1\ne 1+\cot^2 x$, one can easily simplify as follows $$7\sin^2 x\csc^2 x-7\sin^2 x$$ $$=7\sin^2 x(\csc^2 x-1)$$ $$=7\sin^2 x(\cot^2 x)$$ $$=7\sin^2 x\left(\frac{\cos^2 x}{\sin^2 x}\right)$$ $$=\color{red}{7\cos^2x}$$

Harish Chandra Rajpoot
  • 37,450