Question

If cosec theta+cot theta/cosec theta-cot theta=81/49,find cos theta+sin theta/cos theta-sin theta

Answer 1

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Answer 2

Given : $\frac{\csc \theta+\cot \theta}{\csc\theta-\cot\theta}= \frac{81}{49}$

To find : $\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}$

Explanation:

$\frac{\csc \theta+\cot \theta}{\csc\theta-\cot\theta}= \frac{81}{49}\\\\\frac{\frac{1}{\sin \theta}+\frac{\cos\theta}{\sin\theta}}{\frac{1}{\sin \theta}-\frac{\cos\theta}{\sin\theta}}=\frac{81}{49}\\\\\frac{1+\cos\theta}{1-\cos\theta}=\frac{81}{49}\\\\\text{cross multiplying}\\\\49(1+\cos\theta)= 81(1-\cos\theta)\\\\49 + 49 \cos\theta= 81-81\cos\theta\\\\130 \cos\theta = 32\\\\\cos\theta = \frac{32}{130}\\\\\cos\theta = \frac{16}{65}$

$\\\\\sin\theta = \sqrt{1-\cos^2\theta}\\\\\sin\theta = \sqrt{1-(\frac{16}{65})^2}\\\\\text{on solving}\\\\\sin\theta = \frac{63}{65}\\\\now \\\\\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta} = \frac{\frac{63}{65}+\frac{16}{65}}{\frac{63}{65}-\frac{16}{65}}\\\\\frac{\frac{63+16}{65}}{\frac{63-16}{65}}\\\\\frac{79}{47}$

hence , The required value is 79/47

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