Question

Simplify:

$\dfrac{tan\left(x-\dfrac{\pi}{2\right).cos\left(\dfrac{3\pi}{2}+x\right)-sin^3\left(\dfrac{7\pi}{2}-x\right)}}{cos\left(x-\dfrac{\pi}{2}\right).tan\left(\dfrac{3\pi}{2}+x\right)}}$

Answer 1

Answer:

Sin²x

Step-by-step explanation:

tan(x - π/2)  =  -tan(π/2 -x) = -Cotx

cos(3π/2 + x)  = Cos(x - π/2) = Cos(π/2 -x) = Sinx

Sin(7π/2 - x) = Sin(-π/2 - x) = - Sin(π/2 + x) = -Cosx

Sin³(7π/2 - x) = -Cos³x

Cos(x - π/2)  = Cos(π/2 -x) = Sinx

tan(3π/2 + x) = tan(x - π/2) = -tan(π/2 - x)  = - cotx

(-Cotx Sinx  -( -Cos³x) ) / (Sinx (- cotx)

Cotx.Sinx  = Cotx ( cosx . Tanx)  = Cosx

= (-Cosx + Cos³x)/(-cosx)

= 1 - Cos²x

= Sin²x

Answer 2

Answer:

sin²x

Step-by-step explanation:

=> tan(x - π/2) = - tan(π/2 - x) = - cot x

.°. tan(x - π/2) = - cot x

=> cos(3π/2 + x) = cos(x - π/2) = cos(π/2 - x) = sin x

.°. cos(3π/2 + x) = sin x

=> sin(7π/2 - x) = sin(-π/2 - x) = - sin(π/2 - x) = - cos x

.°. sin³(7π/2 - x) = - cos³ x

=> cos(x - π/2) = cos(π/2 - x) = sin x

.°. cos(x - π/2) = sin x

=> tan(3π/2 + x) = tan(x - π/2) = - tan(π/2 - x) = - cot x

.°. tan(3π/2 + x) = - cot x

Now,

=> (- cot x). sin x + cos³x / sin x.(- cot x)

=> - cos x + cos³x / - cos x

=> (1 - cos²x)

=> sin²x ANSWER