Question

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

Answer 1

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

Answer 2

Answer:

Step-by-step explanation:

Sin2a/cos2a+cos2a/sin2a

=sin4a+cos4a/sin2a.cos2a (taking

lcm)

a2+b2=(a+b)2-2ab

=(Sin2a+cos2a)2-

2sin2a.cos2a/sin2a.cos2a

sin2a+cos2a=1

=1-2sin2a.cos2a/sin2a.cos2a

=1/sin2a.cos2a-

2sin2a.cos2a/sin2a.cos2a

=1/sin2a.cos2a-2

=RHS