Question

if sin α *sin β - cos α * cos β + 1= 0 , show that, 1 + cot α * tan β = 0

Answer 1

= sinα*sinβ - cosα*cosβ = -1

multiplying by -1

= cosα*cosβ - sinα*sinβ = 1

= cos(α+β) = 1

= (α + β ) = 0 ..........i)

= 1 + cotα*tanβ

= 1 + (cosα*sinβ /sinα*cosβ)

= sinα*cosβ + cosα*sinβ

= sin(α + β) ......ii)

from equaton i) substituting (α+β) in ii)

= sin(0)

= 0

hope it helped you

Answer 2

Answer:0

Step-by-step explanation: sin alfa sin beta - cos alfa cos beta +1 =0

বা, cos alfa cos beta - sin alfa sin beta =1

বা , cos (alfa + beta ) =1

বা , alfa + beta =0

বা , alfa = - beta

বা , 1 + cot alfa tan beta বা , 1+ cos ( -beta )

বা , 1 - cot beta ×1/ cot beta

বা , 1 -1 =0 proved