ता3% जा5st =cos™ £17 - 85

1.	ता3% जा5st =cos™ £17 - 85
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)}---------(2) {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) {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) {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 {HENCE PROVED} convert sin in terms of tan we get in the formula tan A-tan B/1+ tanA.tanB so we can get the solution

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)}---------(2)

{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)

{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)

{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

{HENCE PROVED}

convert sin in terms of tan we get in the formula tan A-tan B/1+ tanA.tanB

so we can get the solution

Discussion