Question

The value of $\sqrt{3} \left| \frac{\frac{2 \sin(140^\circ)\sec(280^\circ)}{\sec(220^\circ)} + \frac{\sec(340^\circ)}{\csc(20^\circ)}}{\frac{\cot(200^\circ) - \tan(280^\circ)}{\cot(200^\circ)}} \right|$ is ___.

Answer 1

3

Answer 2

To find the value of the given expression, we will simplify the numerator and the denominator separately.

The given expression is $\sqrt{3} \left| \frac{\frac{2 \sin(140^\circ)\sec(280^\circ)}{\sec(220^\circ)} + \frac{\sec(340^\circ)}{\csc(20^\circ)}}{\frac{\cot(200^\circ) - \tan(280^\circ)}{\cot(200^\circ)}} \right|$.

Step 1: Simplify the Numerator

Let the numerator be $N$. $N = \frac{2 \sin(140^\circ)\sec(280^\circ)}{\sec(220^\circ)} + \frac{\sec(340^\circ)}{\csc(20^\circ)}$

First, convert the angles to their equivalent acute angles or related angles in the first quadrant using trigonometric identities:

• $\sin(140^\circ) = \sin(180^\circ - 40^\circ) = \sin(40^\circ)$
• $\sec(280^\circ) = \sec(360^\circ - 80^\circ) = \sec(80^\circ)$ (Since $280^\circ$ is in Quadrant IV, secant is positive)
• $\sec(220^\circ) = \sec(180^\circ + 40^\circ) = -\sec(40^\circ)$ (Since $220^\circ$ is in Quadrant III, secant is negative)
• $\sec(340^\circ) = \sec(360^\circ - 20^\circ) = \sec(20^\circ)$ (Since $340^\circ$ is in Quadrant IV, secant is positive)
• $\csc(20^\circ)$ remains as is.

Substitute these into the numerator expression: $N = \frac{2 \sin(40^\circ)\sec(80^\circ)}{-\sec(40^\circ)} + \frac{\sec(20^\circ)}{\csc(20^\circ)}$

Simplify the first term: $\frac{2 \sin(40^\circ)\sec(80^\circ)}{-\sec(40^\circ)} = -2 \sin(40^\circ) \frac{1/\cos(80^\circ)}{1/\cos(40^\circ)}$ $= -2 \sin(40^\circ) \frac{\cos(40^\circ)}{\cos(80^\circ)}$ $= -\frac{2 \sin(40^\circ)\cos(40^\circ)}{\cos(80^\circ)}$

Using the double angle identity $2 \sin A \cos A = \sin(2A)$: $= -\frac{\sin(2 \times 40^\circ)}{\cos(80^\circ)} = -\frac{\sin(80^\circ)}{\cos(80^\circ)} = -\tan(80^\circ)$

Simplify the second term: $\frac{\sec(20^\circ)}{\csc(20^\circ)} = \frac{1/\cos(20^\circ)}{1/\sin(20^\circ)} = \frac{\sin(20^\circ)}{\cos(20^\circ)} = \tan(20^\circ)$

So, the numerator $N = -\tan(80^\circ) + \tan(20^\circ) = \tan(20^\circ) - \tan(80^\circ)$.

Step 2: Simplify the Denominator

Let the denominator be $D$. $D = \frac{\cot(200^\circ) - \tan(280^\circ)}{\cot(200^\circ)}$

First, convert the angles:

• $\cot(200^\circ) = \cot(180^\circ + 20^\circ) = \cot(20^\circ)$ (Since $200^\circ$ is in Quadrant III, cotangent is positive)
• $\tan(280^\circ) = \tan(360^\circ - 80^\circ) = -\tan(80^\circ)$ (Since $280^\circ$ is in Quadrant IV, tangent is negative)

Substitute these into the denominator expression: $D = \frac{\cot(20^\circ) - (-\tan(80^\circ))}{\cot(20^\circ)}$ $D = \frac{\cot(20^\circ) + \tan(80^\circ)}{\cot(20^\circ)}$ $D = 1 + \frac{\tan(80^\circ)}{\cot(20^\circ)}$

Since $\frac{1}{\cot(20^\circ)} = \tan(20^\circ)$: $D = 1 + \tan(80^\circ)\tan(20^\circ)$

Step 3: Evaluate the fraction inside the absolute value

Let the fraction be $F = \frac{N}{D}$. $F = \frac{\tan(20^\circ) - \tan(80^\circ)}{1 + \tan(20^\circ)\tan(80^\circ)}$

This expression matches the tangent subtraction formula: $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$. Here, $A = 20^\circ$ and $B = 80^\circ$. So, $F = \tan(20^\circ - 80^\circ) = \tan(-60^\circ)$

Since $\tan(-\theta) = -\tan(\theta)$: $F = -\tan(60^\circ) = -\sqrt{3}$

Step 4: Calculate the final value

The problem asks for the value of $\sqrt{3} \left| F \right|$. $\sqrt{3} \left| -\sqrt{3} \right| = \sqrt{3} \times \sqrt{3} = 3$.

The final answer is $\boxed{3}$.

Explanation of the solution:

The given expression is simplified by first converting all trigonometric functions of angles outside the first quadrant to their equivalent forms involving acute angles.

The numerator simplifies to $\tan(20^\circ) - \tan(80^\circ)$ using double angle identities and properties of reciprocal trigonometric functions.

The denominator simplifies to $1 + \tan(80^\circ)\tan(20^\circ)$ by splitting the fraction and using reciprocal identities.

The ratio of the simplified numerator and denominator is recognized as the tangent subtraction formula $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.

This results in $\tan(20^\circ - 80^\circ) = \tan(-60^\circ) = -\sqrt{3}$.

Finally, the absolute value is taken, and the result is multiplied by $\sqrt{3}$, yielding $\sqrt{3} \times |-\sqrt{3}| = \sqrt{3} \times \sqrt{3} = 3$.