Question

if a=cosθ+cotθ and a^2-b^2 =4√(ab) then prove that b=cosθ-cotθ

Answer 1

in other way, question maybe , if a= cosθ + cotθ and we assume b = b=cosθ-cotθ

then prove that a² - b² = 4√ab

solution : LHS = a² - b²

= (cosθ + cotθ)² - (cosθ - cotθ)²

= (cos² + cot²θ + 2cosθ.cotθ) - (cos²θ + cot²θ - 2cosθ.cotθ)

= 4cosθ.cotθ

RHS = 4√(ab)

= 4√{(cosθ + cotθ)(cosθ - cotθ)}

[from algebraic identity, (x - y)(x + y) = x² - y² ]

= 4√{cos²θ - cot²θ}

= 4√{cos²θ - cos²θ/sin²θ}

= 4√{cos²θ(1 - 1/sin²θ)}

= 4cosθ√{1 - cosec²θ}

we know, cosec²x - cot²x = 1

so,1 - cosec²θ = cot²θ

= 4cosθ√{cot²θ}

= 4cosθ. cotθ

here, LHS = RHS

hence, if a = cosθ + cotθ , b = cosθ - cotθ then a² - b² = 4√(ab)

or, if a = cosθ + cotθ and a² - b² = 4√ab then, b = cosθ - cotθ.

[ note : the above method is also a important rule to prove mathematical expressions, if expression are complicated to derive, you can use this method to prove. ]