Question

If cosec θ + cot θ = p, show that cos θ = (P2 -1) / (P2+1)?​

Answer 1

Answer:

Hence Proved, Cos∅ = $(P^{2} -1)/(P^{2} +1)$

Step-by-step explanation:

Given,

Cosec∅ + Cot∅ = P

We know, Cosec∅ = 1/Sin∅  and Cot∅ = Cos∅/Sin∅

Therefore, we can written the above equation as

1/Sin∅ + Cos∅/Sin∅ = P

(1 + Cos∅)/Sin∅ = P

Squaring on both Sides, We get

(1 +Cos∅)2 /$Sin^{2}$∅ = $P^{2}$/1

Now, (Numerator - Denominator)/(Numerator + Denominator), then

{(1 +Cos∅)2 - $Sin^{2}$∅}/{(1 +Cos∅)2 + $Sin^{2}$∅} = ($P^{2}$ - 1)/($P^{2}$ + 1)

{1 +$Cos^{2}$∅ + 2Cos∅ - $Sin^{2}$∅}/{1 + $Cos^{2}$∅ + 2Cos∅ + $Sin^{2}$∅} = ($P^{2}$ - 1)/($P^{2}$ + 1)

{2$Cos^{2}$∅ + 2Cos∅}/{2 + 2Cos∅} = ($P^{2}$ - 1)/($P^{2}$ + 1)

2Cos∅{Cos∅ + 1}/2{Cos∅ + 1} = ($P^{2}$ - 1)/($P^{2}$ + 1)

Thus, Cos∅ = ($P^{2}$ - 1)/($P^{2}$ + 1).

Hence Proved.

Answer 2

given

$\cosec \theta + \cot \theta = p$

Now RHS

$\frac{ {p}^{2} - 1}{ {p}^{2} + 1} \\ \\ = \frac{( {cosec \theta + \cot \theta})^{2} - 1}{( {cosec \theta + cot \theta})^{2} + 1} \\ \\ = \frac{2 {cot}^{2} \theta + 2cosec \theta cot \theta}{2 {cosec}^{2} \theta + 2cosec \theta cot \theta }$

$= \frac{2cot \theta(cot \theta + cosec \theta)}{2coec \theta(cosec \theta + cot \theta)} \\ \\ = \frac{cot \theta}{cosec \theta} \\ \\ = cos \theta$