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))#