Question

If $\frac{\text{cosec}\theta +\cot \theta }{\text{cosec}\theta -\cot \theta }=\frac{81}{49}$, then find $\frac{\cos \theta +\sin \theta }{\sin \theta -\cos \theta }$

• A. $\frac{79}{47}$
• B. $\frac{40}{39}$
• C. $1$
• D. None of these

Answer 1

Answer: (A)

$\frac{79}{47}$

$\frac{\text{cosec}\theta +\cot \theta }{\text{cosec}\theta -\cot \theta }=\frac{81}{49}$
$\Rightarrow 49\text{cosec}\theta +49\cot \theta =81\text{cosec}\theta -81\cot \theta$
$\Rightarrow 49\cot \theta +81\cot \theta =81\text{cosec}\theta -49\text{cosec}\theta$
$\Rightarrow 130\cot \theta =32\text{cosec}\theta$
$\Rightarrow 130\frac{\cos \theta }{\sin \theta }=32\frac{1}{\sin \theta }$
$\Rightarrow 65\cos \theta =16$
$\Rightarrow \cos \theta =\frac{16}{65}$
$P{M}^2=O{P}^2-O{M}^2={65}^2-{16}^2$
$\Rightarrow P{M}^2=(65+16)(65-16)=\left(81\right)\left(49\right)$
$\Rightarrow PM=\sqrt{81\times 49}=9\times 7=63$
$\cos \theta =\frac{OM}{OP}=\frac{16}{65},\sin \theta =\frac{PM}{OP}=\frac{63}{65}$
Required expression $=\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }=\frac{\frac{16}{65}+\frac{63}{65}}{\frac{63}{65}-\frac{16}{65}}=\frac{79}{65}\times \frac{65}{47}=\frac{79}{47}$