Question #d7b02

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Question #d7b02

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Answer 1 · 2017-03-08T06:59:25

See proof below

Explanation:

We need

${cos}^{2} x + {sin}^{2} x = 1$ $cos^2x+sin^2x=1$

$csc x = \frac{1}{sin x}$ $cscx=1/sinx$

$cot x = \frac{cos x}{sin x}$ $cotx=cosx/sinx$

Therefore,

$R H S = cos θ sin θ + {cos}^{3} θ csc θ$ $RHS=cos thetasintheta+cos^3thetacsctheta$

$= cos θ sin θ + \frac{{cos}^{3} θ}{sin θ}$ $=costhetasintheta+cos^3theta/sintheta$

$= \frac{cos θ {sin}^{2} θ + {cos}^{3} θ}{sin θ}$ $=(costhetasin^2theta+cos^3theta)/sintheta$

$= \frac{cos θ ({sin}^{2} θ + {cos}^{2} θ)}{sin θ}$ $=(costheta(sin^2theta+cos^2theta))/sintheta$

$= \frac{cos θ}{sin θ}$ $=costheta/sintheta$

$= cot θ$ $=cottheta$

$= L H S$ $=LHS$

$Q . E . D$ $Q.E.D$

Answer 2 · 2017-03-08T07:01:26

see explanation

Explanation:

Let we take a right hand side to prove left hand side.

$cos θ sin θ + {cos}^{3} θ csc θ = cos θ sin θ + {cos}^{3} θ \cdot \frac{1}{sin θ}$ $cos theta sin theta + cos^3 theta csc theta = cos theta sin theta + cos^3 theta * 1/sin theta$

multiply with

$= \frac{cos θ sin θ sin θ + {cos}^{3} θ}{sin θ}$ $= (cos theta sin theta sin theta+ cos^3 theta) /sin theta$

$= \frac{cos θ {sin}^{2} θ + {cos}^{3} θ}{sin θ}$ $= (cos theta sin^2 theta + cos^3 theta) /sin theta$

$= \frac{cos θ (1 - {cos}^{2} θ) + {cos}^{3} θ}{sin θ}$ $= (cos theta (1-cos^2 theta )+ cos^3 theta) /sin theta$

note: ${sin}^{2} θ = 1 - {cos}^{2} θ$ $sin^2 theta = 1 -cos^2 theta$

$= (cos theta - cancelcos^3 theta + cancelcos^3 theta) /sin theta$

$= cos theta /sin theta$

$= cot theta$

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Question #d7b02

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