Question

prove that.
cos@/sec(90-@) + sin (90-@)/cosec@-1 = 2 cot (90-@)​

Answer 1

Answer:

Step-by-step explanation:

Cosθ/Sec(90 - θ) + 1    + Sin(90 - θ)/Cosecθ + 1

We know that Sin (90-θ) =Cosθ and Sec(90-θ) = Cosecθ

=>  Cosθ/Cosecθ + 1   + Cosθ/Cosecθ + 1  

=> Cosθ(Cosecθ+1) + Cosθ(Cosecθ - 1)/ (Cosecθ + 1)(Cosecθ - 1)

=> Cosθ(Cosecθ + 1 + Cosecθ - 1) / Cosec²θ - 1

=> Cosθ * 2Cosecθ/ Cot²θ     ( ∵Cosec²θ - 1 = Cot²θ)

=> (2Cosθ/Sinθ) / (Cos²θ/Sin²θ)

=> 2CosθSin²θ / Sinθ Cos²θ

=> 2 Tanθ

=> 2Cot(90 - θ)      (∵Tanθ = Cot(90 - θ))

Hence proved.