# Answer on Trigonometry Question for Kewal

Question #5909

Prove that

tan2A = (sec2A+1)√sec²A-1

tan2A = (sec2A+1)√sec²A-1

Expert's answer

(sec(2A)+1)*sqrt(sec^2(A)-1)=(1/cos(2A)+1)*sqrt(1/cos^2(A)-1)=(1+cos(2A))/cos(2A)

)* sqrt( (1-cos^2(A) )/cos^2(A) )=

=2cos^2(A)/(cos(2A) )

*sqrt(sin^2(A)/cos^2(A))=2*(cos^2(A)/cos(2A) )*

sin(A)/cos(A)=2*cos(A)*sin(A)/cos(2A)=sin(2A)/cos(2A)=tan( 2A) and that just we

need we use here sin^2(A)+cos^2(A)=1, sin(2A)=2*sin(A)*cos(A),

1+cos(2A)=2cos^2(A)

)* sqrt( (1-cos^2(A) )/cos^2(A) )=

=2cos^2(A)/(cos(2A) )

*sqrt(sin^2(A)/cos^2(A))=2*(cos^2(A)/cos(2A) )*

sin(A)/cos(A)=2*cos(A)*sin(A)/cos(2A)=sin(2A)/cos(2A)=tan( 2A) and that just we

need we use here sin^2(A)+cos^2(A)=1, sin(2A)=2*sin(A)*cos(A),

1+cos(2A)=2cos^2(A)

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