If sin θ = a cos φ and cos θ = b sin φ

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If sin θ = a cos φ and cos θ = b sin φ , then the value of (a² – 1) cot² φ + (1 – b² ) cot²θ is equal to :

1. a² + b²
  a²
2. a² + b²
  b²
3. a² - b²
  b²
4. a² - b²
  a²

Correct Option: D

(a² - 1)cot²φ + ( 1 - b² )cot² θ

= (a² - 1)	cos²φ	+ ( 1 - b² )	cos²θ
	sin²φ		sin²θ

=	(a² - 1)cos²φ.sin²θ + ( 1 - b² )cos²θ.sin²φ
	sin²φ.sin²θ

=	a²cos²φ.sin²θ - cos²φ.sin²θ + cos²θ.sin²φ - b²cos²θ.sin²φ
	sin²φ.sin²θ

=	sin²θ.sin²θ - cos²φ.sin²θ + cos²θ.sin²φ - cos²θ.cos²θ
	sin²φ.sin²θ

[ ∵ sinθ = bcosφ , cosθ = bsinφ ]

=	sin⁴θ - cos⁴θ - cos²φ.sin²θ + cos²θ.sin²φ
	sin²φ.sin²θ

=	(sin²θ - cos²θ)(sin²θ + cos²θ) - cos²φ.sin²θ + cos²θ.sin²φ
	sin²φ.sin²θ

=	sin²θ - cos²φ.sin²θ - cos²θ + cos²θ.sin²φ
	sin²φ.sin²θ

=	sin²θ( 1 - cos²φ ) - cos²θ( 1 - sin²φ )
	sin²φ.sin²θ

=	sin²θ.sin²φ - cos²θ.cos²φ
	sin²φ.sin²θ

= 1 -	cos²θ.cos²φ	= 1 -	b²
	sin²φ.sin²θ		a²

Required answer =	a² - b²
	a²