Question

For an acute angle $\theta$, if $\sin\theta = \frac{1}{9}$, then value of $\frac{9\csc\theta + 1}{9\csc\theta - 1}$ is

• A. $0$
• B. $\frac{80}{81}$
• C. $1$
• D. $\frac{82}{80}$

Hint:

Always convert complex trigonometric expressions into their basic counterparts.
Since $\csc\theta = \frac{1}{\sin\theta}$, substituting this directly yields $\frac{\frac{9}{\sin\theta} + 1}{\frac{9}{\sin\theta} - 1} = \frac{9 + \sin\theta}{9 - \sin\theta}$, which is extremely quick to compute!

Answer 1

The Correct Option is D

Step 1: Simplify the expression algebraically before substituting.
Write $\csc\theta = \frac{1}{\sin\theta}$ inside the expression: \[ \frac{9\csc\theta+1}{9\csc\theta-1} = \frac{\frac{9}{\sin\theta}+1}{\frac{9}{\sin\theta}-1} \]
Step 2: Multiply numerator and denominator by sin theta to clear the fraction. \[ = \frac{9+\sin\theta}{9-\sin\theta} \] This simplified form avoids computing $\csc\theta$ as a separate step.
Step 3: Substitute sin theta = 1/9 directly. \[ = \frac{9+\frac{1}{9}}{9-\frac{1}{9}} = \frac{\frac{82}{9}}{\frac{80}{9}} = \frac{82}{80} \]
\[ \boxed{\frac{82}{80}} \]