# Show the steps in the identity tg^2 10+ tg 10 tg 70= 1-tg 10 tg 70?

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You may factor out `tan 10^o` to the left side, such that:

`tan 10^o(tan 10^o + tan 70^o) = 1 - tan 10^o*tan 70^o`

Dividing both sides by `1 - tan 10^o*tan 70^o` yields:

`tan 10^o(tan 10^o + tan 70^o)/(1 - tan 10^o*tan 70^o) = 1`

Dividing both sides by `tan 10^o` yields:

`(tan 10^o + tan 70^o)/(1 - tan 10^o*tan 70^o) = 1/tan 10^o`

You should use the following trigonometric identities, such that:

`tan alpha = 1/cot alpha`

`tan alpha = cot(90^o - alpha)`

`tan (alpha + beta) = (tan alpha + tan beta)/(1 - tan alpha*tan beta)`

Using the identities above, yields:

`tan (10^o + 70^o) = cot 10^o => tan 80^o = cot 10^o`

Using the identity `tan alpha = cot(90^o - alpha)` to test if `tan 80^o = cot 10^o` holds, yields:

`tan 80^o = cot(90^o - 80^o) = cot 10^o` valid

**Hence, testing the given identity, yields that `tan^2 10^o + tan 10^o tan 70^o = 1 - tan 10^o tan 70^o` holds.**