Question

sin inverse 3/5-sin inverse 8/17=cos inverse 84/85

Answer 1

$\textbf{Answer}$

We will use following trigonometric formulas in this solution -

•sin^-1(x) - sin^-1(y) = sin^-1{x√(1-y^2) - y√(1-x^2)}---------(1)

•sin^-1(x) = cos^-1{√(1 - x^2)}---------(3)

$\textbf{We have to prove that,}$

sin^-1(3/5) - sin^-1(8/17) = cos^-1(84/85)

LHS = sin^-1(3/5) - sin^-1(8/17)

$\textbf{Using formula (1) in LHS}$

LHS = sin^-1{ 3/5 × √(1 - (8/17)^2 - 8/17 × √1 - (3/5)^2 }

=> LHS = sin^-1{ 3/5 × √(289-64)/17^2 - 8/17 × √8/17 × √(25-9)/25 }

=> LHS = sin^-1{ (3/5 × √(225/289) - 8/17 × √(16/25) }

=> LHS = sin-1 { (3/5 × 15/17) - (8/17 × 4/5) }

=> LHS = sin^-1 (45/85 - 32/85)

=> LHS = sin^-1(13/85)

$\textbf{By using the formula (2) now}$

LHS = cos^-1{√(1 - (13/85)^2)}

=> LHS = cos^-1{√(7225-169)/7225}

=> LHS = cos^-1(√7056/7225)

=> LHS = cos^-1{√(84×84)/(85×85)}

=> LHS = cos^1(84/85) = RHS
$\textbf{HENCE PROVED}$

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