Question

Question #d1324

Answer 1

Answer is below

Explanation:

![https://useruploads.socratic.org/hUdkMrI9Texwzto0F0Wy_IMG_20170823_232257%7E2.jpg)

Enjoy trigonometry

Answer 2

See explanation below.

Explanation:

We have: `frac(1 + sin(x))(1 - sin(x)) = (sec(x) + tan(x))^(2)`

We will begin the proof from the right-hand side of the identity.

First, let's expand the parentheses:

`= sec^(2)(x) + 2 sec(x) tan(x) + tan^(2)(x)`

Then, let's apply the standard trigonometric identities `sec(x) = frac(1)(cos(x))` and `tan(x) = frac(sin(x))(cos(x))`:

`= frac(1)(cos^(2)(x)) + 2 cdot frac(1)(cos(x)) cdot frac(sin(x))(cos(x)) + frac(sin^(2)(x))(cos^(2)(x))`

`= frac(1 + sin^(2)(x))(cos^(2)(x)) + frac(2 sin(x))(cos^(2)(x))`

`= frac(sin^(2)(x) + 2 sin(x) + 1)(cos^(2)(x))`

Let's factorise the numerator of the fraction using the "middle-term break":

`= frac(sin^(2)(x) + sin(x) + sin(x) + 1)(cos^(2)(x))`

`= frac(sin(x)(sin(x) + 1) + 1 (sin(x) + 1))(cos^(2)(x))`

`= frac((sin(x) + 1)(sin(x) + 1))(cos^(2)(x))`

`= frac((sin(x) + 1)^(2))(cos^(2)(x))`

One of the Pythagorean identities is `cos^(2)(x) + sin^(2)(x) = 1`.

We can rearrange it to get:

`Rightarrow cos^(2)(x) = 1 - sin^(2)(x)`

Let's apply this rearranged identity to our proof:

`= frac((1 + sin(x))^(2))(1 - sin^(2)(x))`

Let's factorise the denominator, which is in the form of a difference of two squares:

`= frac((1 + sin(x))^(2))((1 + sin(x))(1 - sin(x)))`

`= frac(1 + sin(x))(1 - sin(x)) " " "` `"Q.E.D."`