Question #0c718

2 Answers
Ratnaker Mehta
Jul 9, 2017

Refer to the Proof in the Explanation.

Explanation:

We know that, #(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca.#

#:. {(1+sinx-cosx)/(1+sinx+cosx)}^2,#

#=(1+sin^2x+cos^2x+2sinx-2sinxcosx-2cosx)/(1+sin^2x+cos^2x+2sinx+2sinxcosx+2cosx),#

#=(2+2sinx-2sinxcosx-2cosx)/(2+2sinx+2sinxcosx+2cosx),#

#={1(1+sinx)-cosx(1+sinx)}/{1(1+sinx)+cosx(1+sinx)},#

#={cancel((1+sinx))(1-cosx)}/{cancel((1+sinx))(1+cosx)},#

#=(1-cosx)/(1+cosx).#

Hence, the Proof.

Tazwar Sikder
Jul 9, 2017

We have: #(frac(1 + sin(x) - cos(x))(1 + sin(x) + cos(x)))^(2)#

#= frac((1 + sin(x) - cos(x))^(2))((1 + sin(x) + cos(x))^(2))#

Expanding the parentheses:

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

#= frac(1 + 2 sin(x) + sin^(2)(x) - 2 cos(x) - 2 sin(x) cos(x) + cos^(2)(x))(1 + 2 sin(x) + sin^(2)(x) + 2 cos(x) + 2 sin(x) cos(x) + cos^(2)(x))#

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

We can rearrange it to get:

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

Let's apply this rearranged identity to our proof:

#= frac(1 + 2 sin(x) + (1 - cos^(2)(x)) - 2 cos(x) - 2 sin(x) cos(x) + cos^(2)(x))(1 + 2 sin(x) + (1 - cos^(2)(x)) + 2 cos(x) + 2 sin(x) cos(x) + cos^(2)(x))#

#= frac(2 + 2 sin(x) - 2 cos(x) - 2 sin(x) cos(x))(2 + 2 sin(x)+ 2 cos(x) + 2 sin(x) cos(x))#

#= frac(2 (1 + sin(x) - cos(x) - sin(x) cos(x)))(2 (1 + sin(x) + cos(x) + sin(x) cos(x))#

#= frac(1 + sin(x) - cos(x) - sin(x) cos(x))(1 + sin(x) + cos(x) + sin(x) cos(x)#

#= frac(sin(x)(1 - cos(x)) + (1 - cos(x)))(sin(x)(1 + cos(x)) + (1 + cos(x)))#

#= frac((1 - cos(x))(sin(x) + 1))((1 + cos(x))(sin(x) + 1))#

#= frac(1 - cos(x))(1 + cos(x)) " " " ""Q.E.D."#

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