Question

2 tan 43
tan 70 °
Cot 20%
Cot 49°​

Answer 1

Question is not getting mate..........

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Answer 2

Answer:

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Step-by-step explanation:

tan(90∘−θ)≡1tanθ , so tan(90∘−θ)tanθ≡1

tan(θ+ϕ)≡tanθ+tanϕ1−tanθtanϕ

So tan70∘=tan(20∘+50∘)=tan20∘+tan50∘1−tan20∘tan50∘

∴tan70∘(1−tan20∘tan50∘)=tan20∘+tan50∘

∴tan70∘−tan70∘tan20∘tan50∘=tan20∘+tan50∘

But tan70∘tan20∘=tan(90∘−20∘)tan20∘=1

∴tan70∘−tan50∘=tan20∘+tan50∘

∴tan70∘=tan20∘+2tan50∘ QED

Note that we can make a general formula, which would look something like:

tanθ≡tan(90∘−θ)+2tan(2θ−90∘)

Or, equivalently:

tanθ≡tanϕ+2tan(θ−ϕ) with θ+ϕ=90∘, θ>ϕ

The question we considered above is this formula with θ=70∘ (ϕ=20∘).