cos a - tan a * cos ((3pi/2) - a) = sec a

1 answer

Let's simplify the left-hand side of the equation:

cos a - tan a * cos ((3pi/2) - a)

Using the identity cos (90 - x) = sin x, we can write the above expression as:

cos a - tan a * sin a

We know that tan a = sin a / cos a, so substituting this in the equation:

cos a - (sin a / cos a) * sin a

Multiplying through by cos a:

cos^2 a - sin^2 a = (cos^2 a) / cos a

Using the Pythagorean identity sin^2 a + cos^2 a = 1, we have:

1 - sin^2 a - sin^2 a = cos a

2sin^2 a = cos a

Recall that sec a = 1 / cos a. Dividing both sides of the equation by cos a:

sec a = 2sin^2 a / cos a

Using the identity sin^2 a = 1 - cos^2 a:

sec a = 2(1 - cos^2 a) / cos a

sec a = (2 - 2cos^2 a) / cos a

sec a = 2/cos a - 2cos a

Using the identity sec a = 1/cos a:

sec a - 2cos a - 2cos a = 1 / cos a

sec a - 4cos a = 1 / cos a

Multiplying through by cos a:

sec a * cos a - 4cos^2 a = 1

But we already know that cos^2 a - sin^2 a = cos a. Therefore:

sec a * cos a - 4(cos^2 a + sin^2 a) = 1

sec a * cos a - 4cos^2 a - 4sin^2 a = 1

Recall that sin^2 a + cos^2 a = 1. So:

sec a * cos a - 4(cos^2 a + 1 - cos^2 a) = 1

sec a * cos a - 4 = 1

sec a * cos a = 5

Now, we are left with sec a * cos a = 5, which is not equal to the original equation of sec a. Therefore, the original equation is not true for all values of 'a'.