Question

if cosec theta + cot theta = m, then find the value of cos theta is-------

Answer 1

Answer:

m²-1/m²+1

Step-by-step explanation:

Given cosec ∅ + sin ∅ = m

=> 1/sin∅ + cos∅/sin∅ = m

=> (1 + cos∅)/sin∅ = m

=> (1 + cos∅) = m sin∅

Squaring on both sides we get,

(1 + cos∅)² = m²sin²∅

But we know that

sin²∅ = 1 - cos²∅

= (1 - cos∅)(1 + cos∅)

cos∅ cannot be -1 , if it is then cosec∅ and cot∅ would be undefined, hence we can cancel out (1 + cos∅) on both sides,

hence we get,

1 + cos∅ = m²(1-cos∅)

=> 1+ cos∅/1-cos∅ = m²

Using componendo and dividendo, we get

1+ cos∅-(1-cos∅)/1+ cos∅+(1-cos∅) = m²-1/m²+1

=> 2cos∅/2 = m²-1/m²+1

=> cos ∅ = m²-1/m²+1.

Answer 2

Solution:

As we know that

$1+ {cot}^{2} \theta = {cosec}^{2} \theta \\ cosec\theta+cot\theta= m\\\\so\\\\\sqrt{1 + {cot}^{2}\theta } + cot \: \theta = m \\ \\ \sqrt{1 + {cot}^{2}\theta } = m - cot \: \theta$

squaring both side

$1 + {cot}^{2} \theta = {m}^{2} + {cot}^{2} \theta - 2m \: cot\theta \\ \\ 1 + 2m \: cot \: \theta = {m}^{2} \\ \\ 2m \: cot \: \theta = {m}^{2} - 1 \\ \\ cot \: \theta = \frac{ {m}^{2} - 1}{2m} = k = \frac{base}{perpendicular} \\ \\ base = ({m}^{2} - 1)k \\ \\ perpendicular = 2mk \\ \\ hypotenuse = k \sqrt{4 {m}^{2} + {m}^{4} + 1 - 2 {m}^{2} } \\ \\ = k\sqrt{{m}^{4} + 1 + 2 {m}^{2}} \\ \\ =k \sqrt{ {( {m}^{2} + 1)}^{2} } \\ \\ hypotenuse= k( {m}^{2} + 1)$

So,
$cos \: \theta = \frac{base}{hypotenuse} \\ \\ cos \: \theta= \frac{ {m}^{2} - 1}{ {m}^{2} + 1}$

Hope it helps you