Question

Evaluate the expression: \[ \frac{\cos 10^\circ + \cos 80^\circ}{\sin 80^\circ - \sin 10^\circ}. \]

• A. \( \tan 35^\circ \)
• B. \( \tan 55^\circ \)
• C. \( \tan 20^\circ \)
• D. \( \tan 70^\circ \)

Hint:

Using sum-to-product identities can greatly simplify expressions involving sums and differences of trigonometric functions. These identities allow us to express complex expressions in a more manageable form.

Answer 1

The Correct Option is B

We are asked to evaluate the expression: \[ \frac{\cos 10^\circ + \cos 80^\circ}{\sin 80^\circ - \sin 10^\circ}. \] Step 1: Use sum-to-product identities.
We can simplify both the numerator and the denominator using the sum-to-product identities for trigonometric functions. The sum-to-product identity for cosines is: \[ \cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right). \] Substituting \( A = 10^\circ \) and \( B = 80^\circ \), we get: \[ \cos 10^\circ + \cos 80^\circ = 2 \cos \left( \frac{10^\circ + 80^\circ}{2} \right) \cos \left( \frac{10^\circ - 80^\circ}{2} \right) = 2 \cos 45^\circ \cos (-35^\circ). \] Since \( \cos(-35^\circ) = \cos 35^\circ \), we have: \[ \cos 10^\circ + \cos 80^\circ = 2 \times \frac{\sqrt{2}}{2} \cos 35^\circ = \sqrt{2} \cos 35^\circ. \] Step 2: Use sum-to-product identity for sines.
The sum-to-product identity for sines is: \[ \sin A - \sin B = 2 \cos \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right). \] Substituting \( A = 80^\circ \) and \( B = 10^\circ \), we get: \[ \sin 80^\circ - \sin 10^\circ = 2 \cos \left( \frac{80^\circ + 10^\circ}{2} \right) \sin \left( \frac{80^\circ - 10^\circ}{2} \right) = 2 \cos 45^\circ \sin 35^\circ. \] Since \( \cos 45^\circ = \frac{\sqrt{2}}{2} \), we have: \[ \sin 80^\circ - \sin 10^\circ = \sqrt{2} \sin 35^\circ. \] Step 3: Simplifying the expression.
Now substitute the simplified expressions for the numerator and denominator into the original equation: \[ \frac{\cos 10^\circ + \cos 80^\circ}{\sin 80^\circ - \sin 10^\circ} = \frac{\sqrt{2} \cos 35^\circ}{\sqrt{2} \sin 35^\circ} = \frac{\cos 35^\circ}{\sin 35^\circ} = \cot 35^\circ. \] Since \( \cot 35^\circ = \tan(90^\circ - 35^\circ) = \tan 55^\circ \), the value of the expression is: \[ \tan 55^\circ. \] Thus, the correct answer is: \[ \boxed{\tan 55^\circ}. \]

Answer 2

We are asked to evaluate the expression:

\[ \frac{\cos 10^\circ + \cos 80^\circ}{\sin 80^\circ - \sin 10^\circ}. \]

Step 1: Apply the sum-to-product identity for cosines.
Using the identity:

\[ \cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right), \]

with \( A = 10^\circ \) and \( B = 80^\circ \), we get:

\[ \cos 10^\circ + \cos 80^\circ = 2 \cos 45^\circ \cos (-35^\circ). \]

Since cosine is an even function, \( \cos (-35^\circ) = \cos 35^\circ \), so:

\[ \cos 10^\circ + \cos 80^\circ = 2 \times \frac{\sqrt{2}}{2} \times \cos 35^\circ = \sqrt{2} \cos 35^\circ. \]

Step 2: Apply the sum-to-product identity for sines.
Using the identity:

\[ \sin A - \sin B = 2 \cos \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right), \]

with \( A = 80^\circ \) and \( B = 10^\circ \), we get:

\[ \sin 80^\circ - \sin 10^\circ = 2 \cos 45^\circ \sin 35^\circ. \]

Since \( \cos 45^\circ = \frac{\sqrt{2}}{2} \), this simplifies to:

\[ \sin 80^\circ - \sin 10^\circ = \sqrt{2} \sin 35^\circ. \]

Step 3: Simplify the overall expression.
Substitute the simplified numerator and denominator:

\[ \frac{\sqrt{2} \cos 35^\circ}{\sqrt{2} \sin 35^\circ} = \frac{\cos 35^\circ}{\sin 35^\circ} = \cot 35^\circ. \]

Recall that:

\[ \cot 35^\circ = \tan (90^\circ - 35^\circ) = \tan 55^\circ. \]

Final answer:

\[ \boxed{\tan 55^\circ}. \]