Prove the following 1- sin A sin (90-1)/cot (90-a)= sin^2 A

Mathematics

1 answer:

Anna [14]3 years ago

7 0

Im going to assume its (90 - A) so we have:-

1 - sinA sin (90 - A) / cot (90 - A)

Convert to sines and cosines:-

= 1 - (sin A cos A) / cos ( 90 - A) / sin ( 90 - A)

= 1 - (sin A cos A) / sin A / cos A

= 1 - cos^2 A

= sin^2 A which is what is required.

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The expression $5 = \sqrt{(x-6^2)+(y - 5)^2}$ can be used to find the coordinate of Point B. Option C is correct.

Given that,
Segment AB has point A located at (6, 5). If the distance from A to B is 5 units, which of the following could be used to calculate the coordinates for point B is to be determined.

<h3>What is the equation?</h3>

The equation is the relationship between variables and represented as y = ax +c is an example of a polynomial equation.

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Now the distance between A (6, 5) and B (x, y) is c = 5 units
So, by the distance formula, the distance between two points is given as,

$C = \sqrt{(x_1-x_2)^2 +(y_1-y_2)^2}$