Question

if cosec x + cot x = k, prove that cos x =k^2-1/k^2+1

Answer 1

given that

cosec x +cot x = k  --------------(1)
we know that 
cosec^2 x -cot^2 x = 1
(cosec x +cot x)(cosec x -cot x) = 1
k (cosec x-cot x) = 1
(cosec x-cot x) = 1/k  ------------------(2)

from (1) & (2)
cosec x+cot x = k
cosec x-cot x = 1/k

2cosec x = k+1/k
2cosec x= k^2+1/k
cosec x = k^2+1/2k
sin x = 2k/k^2+1
we know that 
cos x = √1-sin^2 x
cos x = √1-[2k/k^2+1]^2
cos x = √1-4k^2/(k^2+1)^2
cos x = √(k^2+1)^2-4k^2/(k^2+1)^2
cos x = √(k^2-1)^2/(k^2+1)^2                         {·(a+b)^2-(a-b)^2 = 4ab}
cos x = √[k^2-1/k^2+1]^2 
cos x = k^2-1/k^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