Question

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

Answer 1

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) = sinx

.°.cos(3π/2 + x) = sin x

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

.°.sin³(7π/2 - x) = - cos³ x

=> cos(x - π/2) = cos(π/2 - x) = sinx

.°.cos(x - π/2) = sin x

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

.°.tan(3π/2+x)= - cot x

Now, putting the values.

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

=> - cos x + cos³x / - cos x

=> (1 - cos²x)

=> sin²x ANSWER

Answer 2

ANSWER:

Taking the first term from the question

$\small{ \sf{ tan( x - \dfrac{\pi}{2}) = - tan( {90}^{ \circ} - x) = - cot \: x}}$

.°. - tan(90° - x) = - cot x

Taking the second term from the question

$\small{ \sf{cos( \frac{3\pi}{2} + x) = cos( {270}^{ \circ} + x) = sin \: x}}$

.°. cos(270° + x) = sin x

Taking the third term from the question

$\small{\sf{ - {sin}^{3}( \frac{7\pi}{2} - x) = - [ - {sin}^{3}({90}^{ \circ} - x) ]=cos^{3} \: x }}$

.°. - sin³(90° - x) = cos³ x

Taking the fourth term from the question

$\small{ \sf{cos( x - \frac{\pi}{2} ) = cos( \frac{\pi}{2} - x) = sin \: x}}$

.°. cos(90° - x) = sin x

Taking the fifth term from the question

$\small{ \sf{tan( \frac{3\pi}{2} + x) = tan( {270}^{ \circ} + x) = - cot \: x }}$

.°. tan(270° + x) = - cot x

Substituting the values

We get,

$\small{ \sf {= \frac{ - cot \: x.sin \: x + {cos}^{3}x}{ sin \: x.( - cot \: x)} }} \\ = \small{ \sf{ \frac{ - \frac{cos \: x}{sin \: x}.sin \: x + {cos}^{3} x }{sin \: x \times ( - \frac {cos \: x}{sin \: x} )} }} \\ \small { \sf{ = \frac{ - cos \: x + {cos}^{3} x}{ - cos \: x } }} \\ \small{ \sf{ = 1 + \frac{ - {cos}^{2} x(cos \: x)}{cos \: x} }} \\ \small { \sf{ = 1 - {cos}^{2}x }} \\ \small{ \sf{ = {sin}^{2}x }}$