- #1

chwala

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- Homework Statement
- if ##secθ - cosθ =a##

## cosec θ-sin θ = b## then prove that

##a^2b^2(a^2+b^2+3) =1##

- Relevant Equations
- trigonometry

##a^2= \frac {sin^4θ} {cos^2θ}##

##b^2=\frac {cos^4θ} {sin^2θ}##

therefore

##a^2b^2= sin^2θ. cos^2θ##...1

and

##a^2+b^2+3= \frac {sin^6θ + cos^6θ + 3 sin^2θ cos^2θ} {sin^2θ cos^θ}##...2

multiplying 1 and 2, we get,

##sin^6θ + cos^6θ + 3 sin^2θ cos^2θ## how do i proceed from here?

i think i got it

##sin^6θ + cos^6θ =(sin^2x)^3 + (cos^2x)^3##

using,

##a^3 + b^3 = (a+b)(a^2+b^2-ab)##

= ##(sin^2x + cos^2x)(sin^4x + cos^4x-sin^2x. cos^2x)##

and using,

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

##(a^2+b^2-ab)= (a+b)^2-3ab##

thus,

=## (sin^2x + cos^2x)^2-3sin^2x. cos^2x##

therefore,

##sin^6θ + cos^6θ + 3 sin^2θ cos^2θ## = ## [(sin^2x + cos^2x)^2-3sin^2x. cos^2x]+3 sin^2θ cos^2θ##

=1

Bingo!any alternative method guys?

##b^2=\frac {cos^4θ} {sin^2θ}##

therefore

##a^2b^2= sin^2θ. cos^2θ##...1

and

##a^2+b^2+3= \frac {sin^6θ + cos^6θ + 3 sin^2θ cos^2θ} {sin^2θ cos^θ}##...2

multiplying 1 and 2, we get,

##sin^6θ + cos^6θ + 3 sin^2θ cos^2θ## how do i proceed from here?

i think i got it

##sin^6θ + cos^6θ =(sin^2x)^3 + (cos^2x)^3##

using,

##a^3 + b^3 = (a+b)(a^2+b^2-ab)##

= ##(sin^2x + cos^2x)(sin^4x + cos^4x-sin^2x. cos^2x)##

and using,

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

##(a^2+b^2-ab)= (a+b)^2-3ab##

thus,

=## (sin^2x + cos^2x)^2-3sin^2x. cos^2x##

therefore,

##sin^6θ + cos^6θ + 3 sin^2θ cos^2θ## = ## [(sin^2x + cos^2x)^2-3sin^2x. cos^2x]+3 sin^2θ cos^2θ##

=1

Bingo!any alternative method guys?

Last edited: