Math, asked by Nickk7, 1 year ago

Prove
that.
(Cos2A+Sin2A) /(Cos2A-Sin2A) =(1-Sin2A) /Cos2A​

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Answered by josh07
1

Step-by-step explanation:

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Secondary SchoolMath 13 points

Prove :SIN2A/COS2A+COS2A/SIN2A= 1/SIN2A.COS2A -2

Ask for details Follow Report by Tahira485 14.03.2019

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Answer:

Proved

Step-by-step explanation:

prove :SIN2A/COS2A+COS2A/SIN2A= 1/SIN2A.COS2A -2

Question is

\frac{Sin^2A}{Cos^2A} + \frac{Cos^2A}{Sin^2A} = \frac{1}{Sin^2A Cos^2A} - 2

LHS = \frac{Sin^2A}{Cos^2A} + \frac{Cos^2A}{Sin^2A}

= = \frac{Sin^4A + Cos^4A}{Cos^2A . Sin^2A}\\\\Using\: a^2 + b^2 = (a+b)^2 - 2ab\\\\a = Cos^2A \: \& \:b = Sin^2A\\\\= \frac{(Sin^2A + Cos^2A)^2 - 2Sin^2A Cos^2A}{Cos^2A Sin^2A} \\\\Sin^2A + Cos^2A = 1\\\\= \frac{1 -2Sin^2A Cos^2A}{Cos^2A Sin^2A}

= \frac{1}{Cos^2A Sin^2A} - 2\\\\= RHS

QED

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