Question: Prove the following trigonometric identity: \dfrac{sin 3A}{sinA} - \dfrac{cos 3A}{cosA} = 2.

Solution:


You can also solve the above question by the following process using the formula of compound angles.

Solution:
Given,

LHS
= \dfrac{sin 3A}{sinA} - \dfrac{cos 3A}{cosA}

= \dfrac{sin 3A.cosA - cos3A.sinA}{sinAcosA}

= \dfrac{sin(3A -A)}{sinAcosA}

= \dfrac{sin2A}{sinAcosA}

= \dfrac{2sin2A}{2sinAcosA}

= \dfrac{2sin2A}{sin2A}

= 2
RHS


Related Notes And Solutions:

Link: Introduction To Trigonometry
Link: Values of Trigonometric Ratios
Link: Compound Angles

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