Question #dd425

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Answer 1 · 2017-07-13T18:38:12

$(sin^2xcosx)/(1+sinx)^2.$

The Expression = $(1/(secx+tanx))(sinx/(cscx+1)),$

$={1/(1/cosx+sinx/cosx)}{sinx/(1/sinx+1)},$

$=[1/{(1+sinx)/cosx}][sinx/{(1+sinx)/sinx}],$

$={cosx/(1+sinx)}{sin^2x/(1+sinx)},$

$=(sin^2xcosx)/(1+sinx)^2.$

Answer 2 · 2017-07-13T18:43:30

$frac(sin^(2)(x) cos(x))((1 + sin(x))^(2))$

We have: $(frac(1)(sec(x) + tan(x)))(frac(sin(x))(csc(x) + 1))$

Let's apply three standard trigonometric identities; $sec(x) = frac(1)(cos(x))$ , $csc(x) = frac(1)(sin(x))$ and $tan(x) = frac(sin(x))(cos(x))$ :

$= (frac(1)(frac(1)(cos(x)) + frac(sin(x))(cos(x))))(frac(sin(x))(frac(1)(sin(x)) + 1))$

$= (frac(1)(frac(1 + sin(x))(cos(x))))(frac(sin(x))(frac(1 + sin(x))(sin(x))))$

$= (frac(cos(x))(1 + sin(x)))(frac(sin^(2)(x))(1 + sin(x)))$

$= frac(sin^(2)(x) cos(x))((1 + sin(x))^(2))$