Question

Sin3A-sinA/cos3A+cosA =tanA

Answer 1

Step-by-step explanation:

sin3A-sinA/cos3A+CosA

=2cos2AsinA/2cos2AcosA

=sinA/cosA

=tanA

Answer 2

Trigonometry is a branch of mathematics that studies relationships between the side lengths and the angles of triangles

Step-by-step explanation:

To prove: $\dfrac{\sin 3A- \sin A}{\cos3A+\cos A} =\tan A$

Taking Left Hand Side we have

$\dfrac{\sin 3A- \sin A}{\cos3A+\cos A}$

Now as we know

$\sin x - \sin y = 2 \cos \dfrac{x+y}{2} \sin \dfrac{x-y}{2} \\\\\text {and} \\\\ \cos x - \cos y = 2 \cos \dfrac{x+y}{2} \cos \dfrac{x-y}{2}$

Applying the formula we get

$\dfrac{\sin 3A- \sin A}{\cos3A+\cos A}= \dfrac{2 \cos \dfrac{3A+A}{2} \sin \dfrac{3A-A}{2} }{2 \cos \dfrac{3A+A}{2} \cos \dfrac{3A-A}{2}} =\dfrac{\cos 2A \sin A}{\cos 2A \cos A} = \dfrac{\sin A}{\cos A} =\tan A$

Hence proved

#Learn more

Sin x + sin 3x + sin 5x = 0

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