We know thatTan (x+y)=tanx+tany/1+tanx tanyPut y =xTan(X+x)= tanx+tanx

1.	We know thatTan (x+y)=tanx+tany/1+tanx tanyPut y =xTan(X+x)= tanx+tanx /1-tanx tanyTan2x= 2tanx/1-tan2x
Answer» To prove sin2x = \(\frac{2tanx}{1+tan^2x}\) Let us start from RHS and prove it equal to LHS. RHS = \(\frac{2tanx}{1+tan^2x}\) ⇒\(\frac{2tanx}{sec^2x}\), as 1 + tan²x = sec² x,identity ⇒ \(\frac{2\Big(\frac{sinx}{cosx}\Big)}{\frac{1}{cos^2x}}\) ⇒2sinxcosx = sin2x as per sin2x=2sinxcosx identity

We know thatTan (x+y)=tanx+tany/1+tanx tanyPut y =xTan(X+x)= tanx+tanx /1-tanx tanyTan2x= 2tanx/1-tan2x

Answer»

To prove sin2x = \(\frac{2tanx}{1+tan^2x}\)

Let us start from RHS and prove it equal to LHS.

RHS = \(\frac{2tanx}{1+tan^2x}\)

⇒\(\frac{2tanx}{sec^2x}\), as 1 + tan²x = sec² x,identity

⇒ \(\frac{2\Big(\frac{sinx}{cosx}\Big)}{\frac{1}{cos^2x}}\)

⇒2sinxcosx = sin2x as per sin2x=2sinxcosx identity

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We know thatTan (x+y)=tanx+tany/1+tanx tanyPut y =xTan(X+x)= tanx+tanx /1-tanx tanyTan2x= 2tanx/1-tan2x

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