Question

please help me quickly​

Answer 1

$cotθ \: + \: tanθ \: = \: secθ. \: cosecθ \\ \\ from \: \: lhs \: \: \\ \\ cotθ \: = \frac{1}{tanθ} \: \: \: and \: \: \: \: tanθ \: = \frac{sinθ}{cosθ} \\ \\ \frac{cosθ}{sinθ} + \frac{sinθ}{cosθ} = \frac{ {cos}^{2} θ + {sin}^{2} θ}{sinθ \: . \: cosθ} \\ \\ = \frac{1}{cosθ.sinθ} = \frac{1}{cosθ} \: . \: \frac{1}{sinθ} \\ \\ \boxed{ we \: \: know \: \: that \: \: \frac{1}{cosθ} =secθ \: \: (and) \: \frac{1}{sinθ} = cosecθ} \\ \\ so \: \: \boxed{ secθ \: . \: cosecθ \: \: = \: rhs} \\ \\ \\ \therefore \: \: lhs \: = rhs$

Answer 2

Answer :-

cotθ + tanθ = secθ × cosecθ

Step-by-step explanation :-

cotθ + tanθ = secθ × cosecθ

Considering L.H.S :

cotθ + tanθ

We know :

cotθ = cosθ/sinθ

tanθ = sinθ/cosθ

Taking L.C.M :

(cos²θ + sin²θ)/(sinθ × cosθ)

We know :

cos²θ + sin²θ = 1

So we get :

(1)/(sinθ × cosθ)

1/sinθ × 1/cosθ

We know :

1/sinθ = cosecθ

1/cosθ = secθ

So we get :

∴ secθ × cosecθ = secθ × cosecθ

∴ L.H.S = R.H.S