Question Number 100133 by bemath last updated on 25/Jun/20

If sin^(−1) θ+sin^(−1) β=π θ+β−(2/(θ^2 +β^2 )) = ?

Answered by smridha last updated on 25/Jun/20

sin^(−1) (𝛉)=(π/2)+(π/2)−sin^(−1) (𝛃) sin^(−1) (𝛉)=(𝛑/2)+cos^(−1) (𝛃) 𝛉=sin((𝛑/2)+cos^(−1) (𝛃))=cos(cos^(−1) (𝛃)) so 𝛉=β now 𝛉+β−(2/(𝛉^2 +β^2 ))=2𝛉−(1/(2θ^2 )) or 2𝛃−(1/(2𝛃^2 )).

Commented bybemath last updated on 25/Jun/20

thank you Mister