Question

Prove the following identity true. Show work.

Answer 1

Step-by-step explanation:

Answer is given above

Formulas used

a^2 - b^2 = (a+b)(a-b)

cos^2 t = 1 - sin^2 t

Answer 2

Answer:

The  Pythagorean identity states that

$\sin^2 t + \cos^2 t = 1$

Using that we can rewrite the left denominator as:

$1 - \sin^2 t$

Which can be factored as

$(1 - \sin t)(1+ \sin t)$

The numerator we can expand as:

$(1 - \sin t)(1 - \sin t)$

On the right hand side, let's multply numerator and denominator with (1 - sin t):

The total formula then becomes:

$\dfrac{(1-\sin t)(1-\sin t)}{(1 - \sin t)(1 + \sin t)} = \dfrac{(1-\sin t)(1 - \sin t)}{(1+\sin t)(1 - \sin t)}$

There you go... left and right are equal.