Question

$\frac{\sin 91^{\circ}+\sin 1^{\circ}}{\sin 91^{\circ}-\sin 1^{\circ}}=$

• A. $\tan 46^{\circ}$
• B. $\cot 46^{\circ}$
• C. $\sin 46^{\circ}$
• D. $\cos 46^{\circ}$
• E. 1

Hint:

Trigonometry Tip: The structure $\frac{\sin A + \sin B}{\sin A - \sin B}$ will ALWAYS simplify directly to $\tan(\frac{A+B}{2}) \cot(\frac{A-B}{2})$. Memorizing this saves time!

Answer 1

The Correct Option is A

Concept:
This problem can be simplified drastically by using the sum-to-product trigonometric identities: $$\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$$ $$\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$$

Step 1: Identify the angles A and B.
For both the numerator and the denominator, the angles are: $A = 91^\circ$ $B = 1^\circ$

Step 2: Apply the identity to the numerator.
Use the sum-to-product formula for $\sin A + \sin B$: $$\text{Numerator} = 2 \sin\left(\frac{91^\circ + 1^\circ}{2}\right) \cos\left(\frac{91^\circ - 1^\circ}{2}\right)$$ $$\text{Numerator} = 2 \sin\left(\frac{92^\circ}{2}\right) \cos\left(\frac{90^\circ}{2}\right)$$ $$\text{Numerator} = 2 \sin(46^\circ) \cos(45^\circ)$$

Step 3: Apply the identity to the denominator.
Use the sum-to-product formula for $\sin A - \sin B$: $$\text{Denominator} = 2 \cos\left(\frac{91^\circ + 1^\circ}{2}\right) \sin\left(\frac{91^\circ - 1^\circ}{2}\right)$$ $$\text{Denominator} = 2 \cos\left(\frac{92^\circ}{2}\right) \sin\left(\frac{90^\circ}{2}\right)$$ $$\text{Denominator} = 2 \cos(46^\circ) \sin(45^\circ)$$

Step 4: Construct the simplified fraction.
Place the simplified numerator over the simplified denominator: $$\frac{2 \sin(46^\circ) \cos(45^\circ)}{2 \cos(46^\circ) \sin(45^\circ)}$$ Cancel the constant $2$ from the top and bottom: $$= \left(\frac{\sin(46^\circ)}{\cos(46^\circ)}\right) \cdot \left(\frac{\cos(45^\circ)}{\sin(45^\circ)}\right)$$

Step 5: Evaluate and select the final answer.
Use the identities $\frac{\sin\theta}{\cos\theta} = \tan\theta$ and $\frac{\cos\theta}{\sin\theta} = \cot\theta$: $$= \tan(46^\circ) \cdot \cot(45^\circ)$$ Since $\cot(45^\circ) = 1$: $$= \tan(46^\circ) \cdot 1 = \tan(46^\circ)$$ Hence the correct answer is (A) $\tan 46^{\circ$}.