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

Solution:

LHS
= $\dfrac{sin 2A}{sinA} - \dfrac{cos 2A}{cosA}$

= $\dfrac{sin 2A.cosA - cos 2A.sinA}{sinA.cosA}$

= $\dfrac{sin (2A-A)}{sinA.cosA}$

= $\dfrac{sin A}{sinA.cosA}$

= $\dfrac{1}{cosA}$

= $sec A$
RHS

Related Notes And Solutions:

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

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