3 years ago

Simplify (1+tan^2y)(cos2y-1)

Mathematics

1 answer:

Reil [10]3 years ago

8 0

1+tan^2(y)=1/cos^2(y)
cos 2y -1=cos^2(y) -sin^2(y)-cos^2(y) -sin^2(y)

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) Find the coordinates of the point A' which is the symmetric point

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Let, coordinate of point A' is (x,y).

Since, A' is the symmetric point A(3, 2) with respect to the line 2x + y - 12 = 0.

So, slope of line containing A and A' will be perpendicular to the line 2x + y - 12 = 0 and also their center lies in the line too.

Now, their center is given by :

$C( \dfrac{x+3}{2}, \dfrac{y+2}{2})$

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$\dfrac{y-2}{x-3} \times -2 = -1\\\\2y - 4 = x - 3\\\\2y - x = 1$

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So, $C( \dfrac{2y -1 +3}{2}, \dfrac{y+2}{2})\\\\C( \dfrac{2y + 2}{2}, \dfrac{y+2}{2})$

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