Question

if sin alpha sin beta – cos alpha cos beta +1 =0 then prove that cot alpha tan beta = –1​

Answer 1

Given --->

Sinα Sinβ - Cosα Cosβ + 1 = 0

To prove---> Cotα tanβ = -1

Proof ---> ATQ,

Sinα Sinβ - Cosα Cosβ + 1 = 0

Changing sign of whole equation , we get

=> Cosα Cosβ - Sinα Sinβ - 1 = 0

=> ( Cosα Cosβ - Sinα Sinβ ) = 1

We have a formula,

Cos ( x + y ) = Cosx Cosy - Sinx Siny , applying it here we get,

=> Cos ( α + β ) = Cos0°

=> α + β = 0°

=> β = - α

Now taking ,

LHS = Cotα tanβ

Putting β = - α , in it , we get,

= Cotα tan ( -α )

We know that tan ( -θ ) = -tanθ , applying it here we get,

= Cotα ( - tanα )

We know that , tanθ = 1 / Cotθ , applying it here , we get,

= - Cotα × ( 1 / Cotα )

= - 1 = RHS

Answer 2

Answer:

Step-by-step explanation:

Given :

sinαsinβ−cosαcosβ+1=0

⇒cos(α+β)=−1

α+β=−1

α+β=(2n+1)π

1+cotαtanβ=1+cosαsinα.sinβcosβ

⇒sinαsinβ+cosαsinβsinαcosβ

⇒sin(α+β)sinαcosβ

⇒sin((2n+1)π)sinαcosβ

⇒0

1+cotα+tanβ=0

cotα+tanβ=−1

$S_ta_YH_ap_pY!!!$ ^_^