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What is the value of (1+cot⁡A+tan⁡A)(sin⁡A−cos⁡A)(sin⁡Acos⁡Asin⁡3A−cos⁡3A)(1 + \cot A + \tan A) (\sin A - \cos A) \left(\frac{\sin A \cos A}{\sin^3 A
Question

What is the value of (1+cotA+tanA)(sinAcosA)(sinAcosAsin3Acos3A)(1 + \cot A + \tan A) (\sin A - \cos A) \left(\frac{\sin A \cos A}{\sin^3 A - \cos^3 A}\right)​​

A.

1

B.

-1

C.

0

D.

2

Correct option is A

Given:

(1+cotA+tanA)(sinAcosA)(sinAcosAsin3Acos3A)(1 + \cot A + \tan A) (\sin A - \cos A) \left(\frac{\sin A \cos A}{\sin^3 A - \cos^3 A}\right)

Solution:

=(1+cotA+tanA)(sinAcosA)(sinAcosAsin3Acos3A)=(1+cotA+tanA)(sinAcosA)sinAcosA(sinAcosA)(sin2A+sinAcosA+cos2A)=(1+cosAsinA+sinAcosA)sinAcosAsin2A+sinAcosA+cos2A=sinAcosA(1+cosAsinA+sinAcosA)sin2A+sinAcosA+cos2A=sinAcosA(sin2A+cos2A+sinAcosAsinAcosA)sin2A+sinAcosA+cos2A=sin2A+cos2A+sinAcosAsin2A+sinAcosA+cos2A=1= (1 + \cot A + \tan A) (\sin A - \cos A) \left(\frac{\sin A \cos A}{\sin^3 A - \cos^3 A}\right) \\= \frac{(1 + \cot A + \tan A) (\sin A - \cos A) \sin A \cos A}{(\sin A - \cos A)(\sin^2 A + \sin A \cos A + \cos^2 A)} \\= \frac{(1 + \frac{\cos A}{\sin A} + \frac{\sin A}{\cos A}) \sin A \cos A}{\sin^2 A + \sin A \cos A + \cos^2 A} \\= \frac{\sin A \cos A (1 + \frac{\cos A}{\sin A} + \frac{\sin A}{\cos A})}{\sin^2 A + \sin A \cos A + \cos^2 A} \\= \frac{\sin A \cos A (\frac{\sin^2 A + \cos^2 A + \sin A \cos A}{\sin A \cos A})}{\sin^2 A + \sin A \cos A + \cos^2 A} \\= \frac{\sin^2 A + \cos^2 A + \sin A \cos A}{\sin^2 A + \sin A \cos A + \cos^2 A} \\= 1 \\​​

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