Question

an 50°
don: tan 50°
blem: If A+B=
4
= tan (70° – 20°) = tan 70° - tan 209
1+tan 70°. tan 20°
tan 70° - tan 20º = tan 50°
= tan 50° (1 +tan 70°. tan (90° - 70°))
= tan 50° (1 + tan 70'. cot 70°)
= 2 tan 50°
TT
then prove that
6) (1 + tan A) (1 + tan B) = 2, (ii) (cot A - 1) (cot B - 1) = 2.
1
TT
+B=
4
tan (A + B)= tan
TT
= 1
4
tan A + tan B
=1 = tan A + tan B = 1 - tan A tan B
n​

Answer 1

Answer:

this is the answer

Step-by-step explanation:

There are a couple of useful properties/identities for tangents that we can use here:

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∘ ).