The value of (tan(45ᵒ-α))/(cot(45ᵒ+α))-((cos19ᵒ+sin71ᵒ)(sec \ 19ᵒ+cosec71ᵒ))/(tan \ 12ᵒ tan24ᵒ tan66ᵒtan78ᵒ) is:

The value of (tan(45ᵒ-α))/(cot(45ᵒ+α))-((cos19ᵒ+sin71ᵒ)(sec…

The value of $\frac{tan(45^o-\alpha)}{cot(45^o+\alpha)}-\frac{(cos19^o+sin71^o)(sec \ 19^o+cosec71^o)}{tan \ 12^o tan24^o tan66^otan78^o}$ is:

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Correct answer: A

Explanation

The correct answer is option 1 i.e. -3

$\frac{tan(45^o-\alpha)}{cot(45^o+\alpha)}-\frac{(cos19^o+sin71^o)(sec \ 19^o+cosec71^o)}{tan \ 12^o tan24^o tan66^otan78^o}$

By using the identity,

Tan(90 - x) = Cotx

Cot(90 - x) = Tanx

Sin(90 - x) = Cos x

Cosec (90 - x) = sec x

and similar identites,

⇒ $\frac{cot(90^o-(45^o-\alpha)}{cot(45^o+\alpha)}-\frac{(sin(90^o-19^o)+sin71^o)(cosec(90^o-19^o)+cosec71^o)}{tan \ 12^o tan24^o tan66^otan78^o}$

⇒ $\frac{cot(45^o+\alpha)}{cot(45^o+\alpha)}-\frac{(sin71^o+sin71^o)(cosec71^o+cosec71^o)}{tan \ 12^o cot66^o tan66^ocot12^o}$

⇒ $1-{(2sin71^o)(2cosec71^o)}$

⇒ $1-4$

⇒ $-3$