Math Problem Statement

sec(pi/6)(1+(5tanpi/6-tanpi/6^2)/secpi/6^2)

Solution

Let's break down the given expression step-by-step:

We need to simplify:

sec(π6)(1+5tan(π6)tan(π6)2sec(π6)2)\sec\left(\frac{\pi}{6}\right) \left(1 + \frac{5 \tan\left(\frac{\pi}{6}\right) - \tan\left(\frac{\pi}{6}\right)^2}{\sec\left(\frac{\pi}{6}\right)^2}\right)

Step 1: Evaluate the known trigonometric values.

  • sec(π6)=1cos(π6)=132=23\sec\left(\frac{\pi}{6}\right) = \frac{1}{\cos\left(\frac{\pi}{6}\right)} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}

  • tan(π6)=sin(π6)cos(π6)=1/23/2=13\tan\left(\frac{\pi}{6}\right) = \frac{\sin\left(\frac{\pi}{6}\right)}{\cos\left(\frac{\pi}{6}\right)} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}

  • sec(π6)2=(23)2=43\sec\left(\frac{\pi}{6}\right)^2 = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3}

  • tan(π6)2=(13)2=13\tan\left(\frac{\pi}{6}\right)^2 = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}

Step 2: Substitute the values into the expression.

Substitute the values for sec(π6)\sec\left(\frac{\pi}{6}\right), tan(π6)\tan\left(\frac{\pi}{6}\right), and sec(π6)2\sec\left(\frac{\pi}{6}\right)^2:

23(1+5×131343)\frac{2}{\sqrt{3}} \left(1 + \frac{5 \times \frac{1}{\sqrt{3}} - \frac{1}{3}}{\frac{4}{3}}\right)

Step 3: Simplify the inner expression.

Simplify the expression inside the parentheses:

5×13=535 \times \frac{1}{\sqrt{3}} = \frac{5}{\sqrt{3}}

Now, the expression becomes:

1+5313431 + \frac{\frac{5}{\sqrt{3}} - \frac{1}{3}}{\frac{4}{3}}

Step 4: Combine the terms inside the numerator.

To combine 53\frac{5}{\sqrt{3}} and 13\frac{1}{3}, we need a common denominator. The denominator of 53\frac{5}{\sqrt{3}} is 3\sqrt{3}, so we can rewrite it as:

53=533\frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3}

Now, the expression inside becomes:

1+53313431 + \frac{\frac{5\sqrt{3}}{3} - \frac{1}{3}}{\frac{4}{3}}

Step 5: Simplify further.

Simplify the numerator:

53313=5313\frac{5\sqrt{3}}{3} - \frac{1}{3} = \frac{5\sqrt{3} - 1}{3}

Now the expression is:

1+531343=1+53141 + \frac{\frac{5\sqrt{3} - 1}{3}}{\frac{4}{3}} = 1 + \frac{5\sqrt{3} - 1}{4}

Step 6: Final simplification.

Now multiply everything by 23\frac{2}{\sqrt{3}}:

23(1+5314)\frac{2}{\sqrt{3}} \left( 1 + \frac{5\sqrt{3} - 1}{4} \right)

Distribute the 23\frac{2}{\sqrt{3}}:

23+23×5314\frac{2}{\sqrt{3}} + \frac{2}{\sqrt{3}} \times \frac{5\sqrt{3} - 1}{4}

Simplify the second term:

23×534=104=52\frac{2}{\sqrt{3}} \times \frac{5\sqrt{3}}{4} = \frac{10}{4} = \frac{5}{2}

23×14=243=123\frac{2}{\sqrt{3}} \times \frac{-1}{4} = \frac{-2}{4\sqrt{3}} = \frac{-1}{2\sqrt{3}}

Now the expression becomes:

23+52123\frac{2}{\sqrt{3}} + \frac{5}{2} - \frac{1}{2\sqrt{3}}

Final Answer:

Thus, the simplified expression is:

23+52123\frac{2}{\sqrt{3}} + \frac{5}{2} - \frac{1}{2\sqrt{3}}

This is the most simplified form of the given expression.


Would you like further clarification on any of these steps?

Related questions:

  1. How do you derive sec(π6)\sec\left(\frac{\pi}{6}\right) and tan(π6)\tan\left(\frac{\pi}{6}\right) using unit circle concepts?
  2. What is the geometric meaning of secant and tangent for an angle?
  3. Can you explain the steps for simplifying expressions involving trigonometric functions?
  4. How do you rationalize expressions involving square roots?
  5. How are trigonometric identities useful in simplifying complex expressions?

Tip:

When simplifying trigonometric expressions, it's often helpful to first evaluate all known values for the basic angles, like π6\frac{\pi}{6}, π4\frac{\pi}{4}, and π3\frac{\pi}{3}, to make the algebraic manipulations easier.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Trigonometric Identities
Simplification of Expressions

Formulas

sec(θ) = 1/cos(θ)
tan(θ) = sin(θ)/cos(θ)
sec^2(θ) = 1 + tan^2(θ)

Theorems

Trigonometric Identities

Suitable Grade Level

Grades 10-12