Factor the expression and use the fundamental identities to simplify $7 \sin^2 x \csc^2 x − 7 \sin^2 x$
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 $7\cos(x)$?
$7 \sin^2 x (csc^2 x − 1)$
$=7 \sin^2 x (1 + \cot^2x)$
$=7 \sin^2 x (1 + \frac {\cos^2x}{\sin^2x})$
$=7 \sin^2 x (\frac{\sin^2x}{\sin^2x} + \frac {\cos^2x}{\sin^2x})$
$=7\sin^2x (\frac{\sin^2x+\cos^2x}{\sin^2x})$
$=7\sin^2x (\frac{1}{\sin^2x})$
$=\frac{7\sin^2x}{\sin^2x}$
$=7$
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$\begingroup$$$7\sin^2 x\csc^2 x-7\sin^2 x=7\sin^2 x\frac{1}{\sin^2x}-7\sin^2 x=$$ $$=7-7\sin^2 x=7(1-\sin^2 x)=7\cos^2x$$
$\endgroup$ 1 $\begingroup$Notice, your mistake $\csc^2-1\ne 1+\cot^2 x$, one can easily simplify as follows $$7\sin^2 x\csc^2 x-7\sin^2 x$$ $$=7\sin^2 x(\csc^2 x-1)$$ $$=7\sin^2 x(\cot^2 x)$$ $$=7\sin^2 x\left(\frac{\cos^2 x}{\sin^2 x}\right)$$ $$=\color{red}{7\cos^2x}$$
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