Question

If cosec theta + cot theta = k prove that cos theta = k2-1 / k2 + 1

Answer 1

Answer:

$If \:cosec\theta+cot\theta=k\\then\:cos\theta =\frac{k^{2}-1}{k^{2}+1}$

Step-by-step explanation:

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

$We \: know \: the \\ Trigonometric\:identity:\\cosec^{2}\theta-cot^{2}\theta =1\:---(1)$

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

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

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

/* Subtract equation (2) from equation (1) , we get

$2cot\theta = k-\frac{1}{k}$

$\implies \frac{2cos\theta}{sin\theta}=\frac{k^{2}-1}{k}\:--(3)$

/* Add equations (1) and (2), we get

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

$\implies \frac{1}{sin\theta}=\frac{k^{2}+1}{k}\:---(4)$

$Now,\\Find (3)÷(4),we get$

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

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

Hence, proved.

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Answer 2

Answer:

given

$\cosec \theta + \cot \theta = k$

Now RHS

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

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