# Trigonometry 1

Cosec(sectheta-1)-cottheta(1-costheta)=tantheta-sintheta

Note by Alok Patel
4 years, 7 months ago

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I would like to correct your given which is cosec(sectheta - 1) because this might refer to the actual measure of the side into cosec(theta) times (sec(theta) - 1)... Probably a typo...

Here's the proof:

We will not touch the right-hand side of the equation.

So, (csc(theta)) (sec(theta) - 1) - (cot(theta)) (1-cos(theta)) = tan(theta) - sin(theta)

This is equivalent to:

(1/sin(theta)) ((1/cos(theta)) - 1) - (cos(theta)/sin(theta)) (1 - cos(theta)) = tan(theta) - sin(theta)

Since csc(theta) = 1/sin(theta), sec(theta) = 1/cos(theta), and cot(theta) = cos(theta)/sin(theta):

The left-hand side when expanded becomes:

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

Since sin^2 (theta) + cos^2 (theta) = 1

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

Simplifying the given gives the left-hand side of the equation.

1/sin(theta)cos(theta) - cos(theta)/sin(theta) - sin(theta) = tan(theta) - sin(theta)

This is now equivalent to:

(1 - cos^2 (theta))/sin(theta)cos(theta) - sin(theta) = tan(theta) - sin(theta)

By Pythagorean Identity mentioned earlier:

sin^2 (theta)/sin(theta)cos(theta) - sin (theta) = tan(theta) - sin(theta)

tan(theta) - sin(theta) = tan(theta) - sin(theta)

Hence, identity is proven.

- 4 years, 7 months ago