Question

(cos 3θ/cos θ + sin 3 θ/sin θ ) = 4(2cos^2 θ -1) prove that?​

Answer 1

Step-by-step explanation:

$\frac{cos \: 3 \theta}{cos \: \theta} + \frac{sin \: 3 \theta}{sin \: \theta} = 4(2 {cos}^{2} \theta - 1) \\ LHS = \frac{cos \: 3 \theta}{cos \: \theta} + \frac{sin \: 3 \theta}{sin \: \theta} \: \\ = \frac{cos \: 3 \theta \: sin \: \theta \: + sin \: 3 \theta \: cos \theta}{sin \theta \: cos \: \theta} \\ = \frac{ sin \: 3 \theta \: cos \theta \: + cos \: 3 \theta \: sin \: \theta \: }{sin \theta \: cos \: \theta} \\ = \frac{sin \: (3 \theta \: + \theta)}{sin \theta \: cos \: \theta} \\ = \frac{sin \: 4 \theta}{sin \theta \: cos \: \theta} \: \\ = \frac{2sin \: 2 \theta \: cos \: 2 \theta}{sin \theta \: cos \: \theta} \\ = \frac{2 \times 2sin \: \theta \: cos \: \theta\: cos \: 2 \theta}{sin \theta \: cos \: \theta} \\ = 4 \: cos \: 2 \theta \\ = 4(2 {cos }^{2} \theta - 1) \\ = RHS \\ Thus Proved.$

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