Question

If cot θ + cos θ = m,
and cot θ - cos θ = n,
then prove that,
(m² – n²)² = 16mn

Help... Thanks in advance !​

Answer 1

Step-by-step explanation:

Let theta =x

m=cosecx-sinx=1/(sinx). - sinx/1= (1-sin^2x)/sinx= cos^2x/sinx.

m=cos^2x/sinx………………(1)

n=secx-cosx=1/cosx -cosx=(1-cos^2x)/cosx= sin^2x/cosx.

n= sin^2x/cosx……………….(2)

(m^2.n)^2/3+(n^2.m)^2/3=?

=(cos^4x/sin^2x×sin^2x/cosx)^2/3+(sin^4x/cos^2x×cos^2x/sinx)^2 /3

=(cos^3x)^2/3+(sin^3x)^2/3

= cos^2x+sin^2x

= 1. Answer.

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Answer 2

Let theta =x

m=cosecx-sinx=1/(sinx). - sinx/1= (1-sin^2x)/sinx= cos^2x/sinx.

m=cos^2x/sinx………………(1)

n=secx-cosx=1/cosx -cosx=(1-cos^2x)/cosx= sin^2x/cosx.

n= sin^2x/cosx……………….(2)

(m^2.n)^2/3+(n^2.m)^2/3=?

=(cos^4x/sin^2x×sin^2x/cosx)^2/3+(sin^4x/cos^2x×cos^2x/sinx)^2 /3

=(cos^3x)^2/3+(sin^3x)^2/3

= cos^2x+sin^2x

= 1. Answer.

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