Question: Prove the following trigonometric identity: tan2A +sin2A = $\dfrac{4tanA}{1-tan⁴A}$
Solution:
Given,
LHS
= tan 2A + sin 2A
= $\dfrac{2 tanA}{1 -tan²A} + \dfrac{2 tanA}{1 +tan²A}$
= $\dfrac{2 tanA(1 +tan²A) + 2tanA (1 -tan²A)}{(1 -tan²A)(1 +tan²A)}$
= $\dfrac{2tanA (1 +tan²A +1 -tan²A}{1 -tan⁴A}$
= $\dfrac{2tanA (2)}{1 -tan⁴A}$
= $\dfrac{4tanA}{1 -tan⁴A}$
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