Answer to Question #162841 in Calculus for Aldryan venth

Question #162841

Sec²(A/2)= 2-2 cos A/ sin² A


1
Expert's answer
2021-02-24T07:25:31-0500

Use the trigonometric identity "1-cos\\theta=2sin^2\\theta" to write the numerator as,


"2-2cosA=2(1-cosA)=4sin^2(\\frac{A}{2})"


Use trigonometric identity "sin\\theta=2sin({\\frac{\\theta}{2})}cos({\\frac{\\theta}{2})}" to write the denominator as,


"sinA=2sin({\\frac{A}{2})}cos({\\frac{A}{2})}"


From Right-Hand side,


"\\frac{2-2cosA}{sin^2A}=\\frac{4sin^2(\\frac{A}{2})}{(2sin\\frac{A}{2}cos\\frac{A}{2})^2}"


"=\\frac{4sin^2(\\frac{A}{2})}{4sin^2\\frac{A}{2}cos^2\\frac{A}{2}}"


The common term "4sin^2(\\frac{A}{2})" is cancel out from numerator and denominator.


"=\\frac{1}{cos^2(\\frac{A}{2})}"


"=sec^2(\\frac{A}{2})=" Left-Hand side


Therefore, "\\frac{2-2cosA}{sin^2A}=" "sec^2(\\frac{A}{2})" , hence proved.

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