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How do you prove #(tan3t-tant)/(1+tan3t*tant)=(2tant)/(1-tan^2t)#?

Answer 1

Victoria Adebayo

#LHS=(tan3t-tant)/(1+tan3t*tant)#

#=((sin3t)/(cos3t)-(sint)/(cost))/(1+(sin3t)/(cos3t)*(sint)/(cost))#

#=(sin3tcost-cos3tsint )/(cos3tcost+sin3tsint)#

#=(sin2t)/(cos2t)#

#=tan2t#

#=(2tant)/(1-tan^2t)=RHS#

Proved

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Answer 2

Caroline Adams

To prove (tan3t - tan t) / (1 + tan3t * tan t) = (2tan t) / (1 - tan^2 t), we'll start with the left-hand side (LHS) and manipulate it step by step using trigonometric identities.

LHS: = (tan3t - tan t) / (1 + tan3t * tan t) = (tan(2t + t) - tan t) / (1 + tan(2t + t) * tan t) = [(tan2t + tan t) / (1 - tan2t * tan t)] / (1 + tan2t * tan t) = (tan2t + tan t) / (1 - tan^2 t) = 2tan t / (1 - tan^2 t)

This matches the right-hand side (RHS), thus proving the identity.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.