Question

if cosec theta+cot theta =k then prove that cos theta =ksquare-1/ksquare+1

Answer 1

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Answer 2

Answer:

Given,

$cosec\theta + cot \theta = k$  ------(2)

By the trigonometry identity,   cosec² A - cot² A = 1,

$cosec^2 \theta - cot^2 \theta = 1$

$(cosec \theta - cot \theta)(cosec \theta + cot \theta) = 1$

$(cosec \theta - cot \theta).k = 1$

$(cosec \theta - cot \theta) = \frac{1}{k}$ ------(2),

Equation (2) + Equation (1),

We get,

$2cosec \theta= k + \frac{1}{k}=\frac{k^2+1}{k}$

$\implies cosec \theta = \frac{k^2+1}{2k}$

$\implies \frac{1}{sin\theta}=\frac{k^2+1}{2k}$

$\implies sin \theta = \frac{2k}{k^2+1}$

Now, we know that, cos A = √(1-sin²A)

$\implies cos A = \sqrt{1-(\frac{2k}{(k^2+1})^2}$

$\implies cos A = \sqrt{1-\frac{4k^2}{(k^2+1)^2}}$

$= \sqrt{(k^2+1)^2-\frac{4k^2}{(k^2+1)^2}}$

$= \sqrt{\frac{k^4+2k^2+1-4k^2}{(k^2+1})^2}$

$= \sqrt{\frac{k^4-2k^2+1}{(k^2+1)^2}$

$= \sqrt{\frac{(k^2-1)^2}{(k^2+1)^2}}$

$=\frac{(k^2-1)}{(k^2+1)}$

Hence, proved.....