Question

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Answer 1

go thru both once

Answer 2

but why are you dividing both top and bottom by sin(x)cos(x)?

Answer 3

ok..

Answer 4

cuz we want 1 in the top

Answer 5

idk the actual q... just looking at that work, it appears like we want to have 1 in the top.....

Answer 6

Ok,mind if i work it out again?

Answer 7

ha

Answer 8

problem: \[\large (\sec(x)-\cos(x))(\csc(x)-\sin(x)) = \frac{1}{\tan(x) +\cot (x)}\]
\[\large \left( \frac{1}{\cos(x)}- \cos(x) \right ) \left( \frac{1}{\sin(x)} -\sin(x)\right)\]\[\large \left(\frac{1-\cos^2(x)}{\cos(x)}\right) \left(\frac{1-\sin^2(x)}{\sin(x)}\right)\]\[\large \left(\frac{\sin^2(x)}{\cos(x)}\right)\left(\frac{\cos^2(x)}{\sin(x)}\right)\]\[= \large \frac{\sin(x)\cos(x)}{1}\]\[=\large \frac{\sin(x)\cos(x)}{\sin^2(x)+\cos^2(x)}\]\[\large \frac{1}{\frac{\sin(x)\cos(x)}{\sin^2(x)+\cos^2(x)}}\]
\[\huge \frac{1}{\frac{1}{\frac{\sin(x)\cos(x)}{\sin^2(x)+\cos^2(x)}}}\]\[\large \frac{1}{\frac{\sin^2(x)+\cos^2(x)}{\sin(x)\cos(x)}}\]\[\large \frac{1}{\tan(x)+\cot(x)}\]

Answer 9

got it now... its an identity :) nice...

Answer 10

Oh I see. if two sides in a triangle are congruent, then the 3rd side must be congruent as well :) Cool.

Answer 11

Oh I just think of a circle..

Answer 12

:)

Answer 13

For these triangles to be congruent by SAS, the included also has to be congruent. else they can open up to anything and the triangles cannot be congruent/similar.