Question

Cosec theta +cot theta=k then prove that cos theta =$\frac{k^{2}-1}{k^{2}+1}$

Answer 1

Step-by-step explanation:

$Cosec∅+cot∅=k→(1)$

$cosec²∅-cot²∅=1$

$(cosec∅+cot∅)(cosec∅-cot∅)=1$

$k(cosec∅-cot∅)=1$

$cosec∅-cot∅=\frac{1}{k}→(2)$

add equation (1) & (2),

$cosec∅+cot∅=k$

$cosec∅-cot∅=\frac{1}{k}$

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$2cot∅=k-\frac{1}{k}$

$2cot∅=\frac{k²-1}{k}$

$cot∅=\frac{k²-1}{2k}$

Then,

$cosec∅+cot∅=k$

$cosec∅-cot∅=\frac{1}{k}$

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$2cosec∅=k+\frac{1}{k}$

$2cosec∅=\frac{k²+1}{k}$

$cosec∅=\frac{k²+1}{2k}$

thus,

$cosec∅=sin∅$

$sin∅={k²+1}{2k}$

then,

$cot∅=\frac{cos∅}{sin∅}$

$\frac{k²-1}{2k}=\frac{\frac{cos∅}{k²+1}}{2k}$

$cos∅=\frac{k²-1}{2k}-\frac{2k}{k²+1}$

Here, both 2k gets cancel

$\frac{k²-1}{k²+1}$

Hence proved

Answer 2

Given:-

❥ cosec θ + cot θ = k

To prove :-

❥ cos θ = k² - 1 / k² + 1

Proof:-

We know ,

❥ cosec θ = 1/sinθ  & cotθ=cosθ/sinθ

Given ,

➥ cosec θ + cot θ= k

➥ 1/sin θ+cos θ/sin θ=k

➥1 + cos θ /sin θ = k

➥1 + cos θ = k sin θ

Squaring on both sides :

➥(1+ cos θ)² = k²sin²θ

➥(1+cos θ)² = k²(1- cos²θ)

➥(1+cos θ)(1+cos θ)=k²(1+cosθ)(1-cosθ)

➥1 + cos θ = k²( 1- cos θ)

➥1 + cos θ = k²- k²cosθ

➥ 1 + ( 1+k² )cos  = k²

➥ (1+k²)cos θ = k² - 1

➥cos θ= k²-1 / k²+1

∴ $\boxed{\pink{\sf cos\theta =k^{2} -1 / k^2 +1 }}$

Hence  proved !

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