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Answer 1 · 2017-08-23T17:57:45

Answer is below

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Enjoy trigonometry

Answer 2 · 2017-08-24T16:18:44

See explanation below.

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."#