Question

if cosec + cot =p, then prove that cos = p2- 1/p2 + 1

Answer 1

We know that
cosec^2 - cot^2 = 1
(cosec - cot)×(cosec+cot) = 1
cosec - cot = (1/p)

cosec + cot = p

2cosec = p + 1/p
2cosec = (p^2 + 1)/p
cosec = (p^2 + 1)/2p
sin = 2p/(p^2 + 1)
sin^2 = 4p^2/((p^2 + 1)^2)
1 - sin^2 = cos^2 = ((p^2 - 1)^2)/((p^2 + 1)^2)
cos = (p^2 - 1)/(p^2 + 1)

Hence Proved

Answer 2

Answer:

Step-by-step explanation:

Cosec A + Cot A = P

= 1/Sin A + Cos A / Sin A = P

= 1+Cos A/Sin A = p

=> SQUARING ON BOTH SIDES

= (1+COS A)²/(SINA)²= P²

= (1+ Cos A)² / (Sin A)² = P²

= (1+cos A)² = (p²)[(Sin A)²]

= (1+ cos A) ² = (p²) [(1-cos²A)]

= (1+ cos A) ² = (p²) [ (1+cosA)(1-cos A) ]

= (1+cos A)² ÷ (1+ cos A) = (p²)[(1-cos A)]

= 1+cos A = (p²)[1-cos A]

= 1+cos A ÷ 1-cos A = p²

= Here, Using (a+b/a-b=c+d/c-d). This is known as Componendo and dividendo

According to the Question statement!

1+cos A ÷ 1-cos A = p²

Then,

(1+cos A) + (1 - cos A ) ÷ (1+ Sin A - (1-sin A) = p²+1 / p²-1

= 2/2cos = p²+1/p²-1

= 1/cos = p²+1/p²-1

= Sec = p²+1 /p²-1

We know that Cos A = 1/sec A

Then,

Cos A = p²-1 /p²+1