Prove that tanθ(1+sec2θ)=tan2θ

Steps to prove tanθ(1+sec2θ)=tan2θ

Take L.H.S tanθ(1+sec2θ)

=tanθ(1+sec2θ)

=tanθ(1+1/cos2θ)

=tanθ{1+(1+tan^2θ)/(1-tan^2θ)}

=tanθ(1-tan^2θ+1+tan^2θ)/(1-tan^2θ)

=tanθ(2/1-tan^2θ)

=2tanθ/1-tan^2θ

=tan2θ

=R.H.S

Hence tanθ(1+sec2θ)=tan2θ is proved.

Detail Information:-

Similar questions for you:- 

• Prove that tan^2A-tan^2B=sin(A+B).sin(A-B)/cos^2A.cos^2B
• Prove that cotA-tanA=2cot2A
• Prove that tan3A-tan2A-tanA=tan3A.tan2A.tanA
• Prove that tanA/2=√(1-cosA)/(1+cosA)