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Temka [501]
3 years ago
11

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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