Last Activity: 12 Years ago
Hi Pradhan,
Start this problem from the right end of the LHS
2tan2A+4cot4A = 2tan2A+4/tan4A = 2tan2A + 2(1-tan22A)/(tan2A) = 2/tan2A
So tanA+2tan2A+4cot4A = tanA + 2/tan2A = tanA + (1-tan2A)/tanA = 1/tanA = cotA.
Best Regards,
Ashwin (IIT Madras).
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Last Activity: 2 Year ago(s)