Answer to Question #5909 in Trigonometry for Kewal

Question #5909
Prove that
tan2A = (sec2A+1)√sec²A-1
1
Expert's answer
2012-01-06T11:36:39-0500
(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)

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS