prove it tan a + cot a + tan 3a +cot 3a = 8 cos^2 2a/ sin 6a . please help me right now! tq

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tan a + cot a + tan 3a +cot 3a = 8 cos^2 2a/ sin 6a

LHS=tan a+cot a + tan (3a) +cot (3a)

=tan a + 1/ (tan a )+ tan (3a) + 1/ (tan (3a) )

=(tan^2(a)+1)/(tan(a))+ (tan^2(3a)+1)/(tan(3a))

=sec^2(a)/tan(a) +sec^2(3a)/tan(3a)

=1/(cos(a)sin(a))+1/(cos(3a)sin(3a))

=(cos(3a)sin(3a)+cos(a)sin(a))/(cos(a)sin(a)cos(3a)sin(3a))

=(2cos(3a)sin(3a)+2cos(a)sin(a))/((2cos(a)sin(a))cos(3a)sin(3a))

=(sin(6a)+sin(2a))/(sin(2a)cos(3a)sin(3a))

=2(2sin(4a)cos(2a))/(sin(2a) 2 cos(3a)sin(3a))

=(4sin(4a)cos(2a))/(sin(2a)sin(6a))

=(4 .2 sin(2a) cos(2a) cos(2a))/(sin(2a) sin(6a))

=(8cos^2 (2a))/(sin(6a))

=RHS

Hence proved.

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