# Show the equality tg 9 degrees + ctg 9 degrees = 4+ctg 27 degees + tg 27 degrees?

You need to re-arrange the terms, such that:

`tan 9^o - tan 27^o + cot 9^o - cot 27^o = 4`

`(sin 9^o)/(cos 9^o) - (sin 27^o)/(cos 27^o) + (cos 9^o)/(sin 9^o) - (cos 27^o)/(sin 27^o) = 4`

`(sin 9^o*cos 27^o - sin 27^o*cos 9^o)/(cos 9^o*cos 27^o) + (sin 27^o*cos 9^o - sin 9^o*cos 27^o)/(sin 9^o*sin 27^o) = 4`

Using the formula `sin(a - b) = sin a*cos b - sin b*cos a` yields:

`(sin (9^o - 27^o))/(cos 9^o*cos 27^o) + (sin (27^o - 9^o))/(sin 9^o*sin 27^o) = 4`

`(sin(-18^o))/(cos 9^o*cos 27^o) + (sin 18^o)/(sin 9^o*sin 27^o) = 4`

Using the formula `sin(-a) = -sin a` yields:

`(sin 18^o)/(sin 9^o*sin 27^o) - (sin 18^o)/(cos 9^o*cos 27^o) = 4`

Factoring out `sin 18^o` yields:

`sin 18^o*(cos 9^o*cos 27^o - sin 9^o*sin 27^o)/(sin 9^o*cos 9^o* sin 27^o*cos 27^o) = 4`

Using the formula `cos (a + b) = cos a*cos b - sin a*sin b` yields:

`sin 18^o*(cos (9^o + 27^o))/(sin 9^o*cos 9^o* sin 27^o*cos 27^o) = 4`

Multilplying by 4 the denominator, you can use then the double angle identity `2sin a*cos a = sin 2a` , such that:

`4*sin 18^o*(cos (9^o + 27^o))/(2*sin 9^o*cos 9^o*2sin 27^o*cos 27^o) = 4`

`4*sin 18^o*(cos (9^o + 27^o))/(sin (2*9^o)*sin (2*27^o)) = 4`

Reducing duplicate terms yields:

`sin 18^o*(cos (36^o))/(sin (18^o)*sin (54^o)) = 1`

Reducing duplicate terms yields:

`(cos (36^o))/(sin (54^o)) = 1`

You need to use the following trigonometric identity, such that:

`sin(90^o - a) = cos a`

Reasoning by analogy, yields:

`sin (90^o - 36^o) = cos 36^o`

`sin 54^o = cos 36^o`

Replacing `sin 54^o` for `cos 36^o` yields:

`(cos (36^o))/(sin (54^o)) = (sin 54^o)/(sin 54^o) = 1`

Hence, testing if the given expression holds, using all indicated the trigonometric identities, yields that `tan 9^o - tan 27^o + cot 9^o - cot 27^o = 4` is valid.

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