Question

(Sec(theta)*sin(theta))/(tan(theta)+cot(theta))=sin^2(theta)

Answer 1

I am going to use x for theta
(Secx*sinx)/(tanx+cotx)=sin^2 x

LS = (1/cosx * sinx)/(sinx/cosx + cosx/sinx)
= (sinx/cosx)/((sin^2 x + cos^2 x)/(sinxcosx))
= (sinx/cosx)(sinxcosx/1)
= can you see it happening?

Answer 2

no,

Answer 3

Really?
Just cancel the cosx , top and bottom

Answer 4

sec θ ∙ sin θ / ( tan θ + cot θ) =

( 1 / cos θ ) ∙ sin (θ) / ( sin θ / cos θ + cos θ / sin θ )n=

( 1 / cos θ ) ∙ sin θ / [ ( sin θ ∙ sin θ + cos θ ∙ cos θ ) / ( cosθ ∙ sin θ ) ] =

( 1 / cos θ ) ∙ sin θ / [ ( sin² θ + cos² θ ) / ( cosθ ∙ sin θ ) ] =

( 1 / cos θ ) ∙ sin θ / [ 1 / ( cosθ ∙ sin θ ) ] =

( 1 / cos θ ) ∙ sin θ ∙ 1 / [ 1 / ( cosθ ∙ sin θ ) ] =

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Remark:

1 / ( 1 / x ) = x

that's why

1 / [ 1 / ( cosθ ∙ sin θ ) ] = cosθ ∙ sin θ
______________________________

( 1 / cos θ ) ∙ sin θ ∙ cosθ ∙ sin θ =

( 1 / cos θ ) ∙ cosθ ∙ sin θ ∙ sin θ =

( cos θ / cos θ ) ∙ sin θ ∙ sin θ =

1 ∙ sin² θ = sin² θ