Question

If cosec theta+ cot theta=k, then prove cos theta= k^2-1/k^2+1​

Answer 1

Answer:

your answer attached in the photo

Answer 2

Answer:

$\huge\underline{ \underline{ \red{your \: \: answer}}}$

Step-by-step explanation:

$\sf\red{cosec \theta - cot \theta == k} \\ \\ \sf\pink{to \:prove = = \frac{ {k}^{2} - 1}{ {k}^{2} + 1 }==cos \theta} \\ \\ \sf\pink{LHS = \frac{ {k}^{2} - 1 }{ {k}^{2} + 1 } }\\ \\ \sf\blue{ = =>\frac{(cosec\theta-cot\theta)^2-1}{(cosec\theta-cot\theta)^2+1}} \\\\ \sf\green{= = > \frac{{cosec \: \theta}^{2} +{ co t \: \theta}^{2} + cosec \: \theta \: cot\theta- 1}{ {cosec \: \theta}^{2} + {cot\theta} ^{2} + cosec\theta \: cot\theta + 1}} \\ \\ \sf\purple{ = => \frac{ 2{cot\theta} ^{2} + 2cosec\theta \: cot\theta}{2{cosec \: \theta}^{2} + 2cosec\theta \: cot\theta} } \\ \\ \sf\pink{ = => \frac{2({cot \theta} ^{2} + cosec\theta \: cot\theta)}{2({cosec \: \theta}^{2} + cosec\theta \: cot\theta)}} \\ \\ \sf\orange{ == > \frac{cot \theta(cot \theta + cosec \theta)}{cosec \theta(cosec \theta + cot \theta)} } \\ \\ \sf\red{== > \frac{cot \theta}{cosec \theta} } \\ \\ \sf\pink{= = > \frac{ \frac{ cos \theta}{sin \theta} }{ \frac{1}{sin \theta} } } \\ \sf\green{== > \frac{cos \theta}{sin \theta} \times \frac{sin \theta}{1} } \\ \\ \sf\blue{== > cos \theta} \\ \\ \sf\purple{RHS}$

$1 + {cot}^{2} \theta = {cosec}^{2} \\ - 1 + {cosec}^{2} \theta = { cot}^{2} \theta \\ 1 - {cosec}^{2} \theta = {cot}^{2} \theta$

$\large\boxed{ \fcolorbox{violet}{yellow}{hope \: it \: help \: you}}$