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# Prove the following: (cos 4x + cos 3x + cos 2x) / (sin 4x + sin 3x + sin 2x) = cot 3x

**Solution:**

LHS = (cos 4x + cos 3x + cos 2x) / (sin 4x + sin 3x + sin 2x)

= ([cos 4x + cos 2x] + cos 3x) / ([sin 4x + sin 2x] + sin 3x)

Since cos A + cos B = 2cos [(A + B) / 2] cos [(A - B) / 2] and sin A + sin B = 2sin [(A + B) / 2] cos [(A - B) / 2],

= {[2cos (4x + 2x)/2 cos (4x - 2x)/2] + cos 3x} / {[2sin (4x + 2x)/2 cos (4x - 2x)/2] + sin 3x}

= (2cos 3x cos x + cos 3x) / (2sin 3x sin x + sin 3x)

= [cos 3x(2cos x + 1)] / [sin 3x(2cos x + 1)]

= cos 3x / sin 3x

= cot 3x

= RHS

NCERT Solutions Class 11 Maths Chapter 3 Exercise 3.3 Question 21

## Prove the following: (cos 4x + cos 3x + cos 2x) / (sin 4x + sin 3x + sin 2x) = cot 3x

**Summary:**

We got, (cos 4x + cos 3x + cos 2x) / (sin 4x + sin 3x + sin 2x) = cot 3x. Hence Proved

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