# What is tg 130^0+tg 50^o=?

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### 2 Answers

You should use the following trigonometric identity, such that:

`tan alpha = (sin alpha)/(cos alpha)`

Reasoning by analogy, yields:

`tan 130^o = (sin 130^o)/(cos 130^o)`

`tan 50^o = (sin 50^o)/(cos 50^o)`

Replacing `(sin 130^o)/(cos 130^o)` for `tan 130^o` and `(sin 50^o)/(cos 50^o)` for tan `50^o` , yields:

`tan 130^o + tan 50^o = (sin 130^o)/(cos 130^o) + (sin 50^o)/(cos 50^o)`

You need to bring the terms to a common denominator, such that:

`tan 130^o + tan 50^o = (sin 130^ocos 50^o + sin 50^ocos 130^o)/(cos 130^ocos 50^o)`

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

`sin a*cos b + sin b*cos a = sin(a + b)`

Reasoning by analogy, yields:

`sin 130^ocos 50^o + sin 50^ocos 130^o = sin(130^o + 50^o)`

`sin 130^ocos 50^o + sin 50^ocos 130^o = sin(180^o)`

Since `sin(180^o) = 0` yields:

`tan 130^o + tan 50^o = (sin(180^o))/(cos 130^ocos 50^o)`

`tan 130^o + tan 50^o = 0/(cos 130^ocos 50^o)`

`tan 130^o + tan 50^o = 0`

**Hence, evaluating the summation `tan 130^o + tan 50^o` , yields `tan 130^o + tan 50^o = 0` .**

**Sources:**

`tan(130^)=tan(90^o +40^o)=-cot(40^o)`

`` `tan(50^o)=tan(90^o-40^o)=cot(40^o)`

`Thus`

`tan(150^o)+tan(50^o)=-cot(40^o)+cot(40^o)=0`

Ans= 0