If 0- - 90o and tan- -3 and tan- -2 then value of sin 2- is-

Given tan (α+β)=3,tan (α β)=2 tan 2 α=tan ((α+β)+(α β))=tan (α+β)+tan (α β)1+tan (α+β)tan (α β)=3+21 (3)(2)= 1 As 0≤α≤ 90^∘ 0≤ 2α≤ 180^∘ ...

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Complementary Trigonometric Ratios

If 0≤α, β≤ ...

Question

If $0 \leq α, β \leq 90^{o}$ and $tan (α + β) = 3$ and $tan (α - β) = 2$ then value of $sin 2 α$ is?

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0 \leq α, β \leq 90^{\circ}

t a n (α + β) = 3

t a n (α - β) = 2

s i n 2 α

tan (α + β) + tan (α - β) = \frac{sin 2 α}{1 - {sin}^{2} α - {sin}^{2} β}

sin 2 α = 4 sin 2 β

\frac{5 tan (α - β)}{tan (α + β)} =

\frac{t a n (α + β - γ)}{t a n (α - β + γ)} = \frac{t a n γ}{t a n β}

s i n (β - γ) = 0

s i n 2 α + s i n 2 β + s i n 2 γ = 0

α = 30^{\circ}

β = 60^{\circ}

\frac{sin α + sec 2 α + tan (α + 15^{\circ})}{tan β + cot (\frac{β}{2} + 15^{\circ}) + tan α}

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