Question

If cot B = $\sqrt{3}$, find the value of $\frac{cos^{2} B-sin^{2}B }{2cosB . sin B}$

Answer 1

Answer:

Since we know that sin B = 2sin(B/2)cos(B/2) and 1 – cos B = 2 sin2(B/2), put all these things

in the required equation

⇒ cot A = (2sin(B/2)cos(B/2))/( 2 sin2(B/2)) = cot (B/2)

⇒ cot (2A) = cot (2 × B/2) = cot (B)

∴ The simplified value is cot(B)

Step-by-step explanation:

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