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Trigonometry: Competency Practice

Pythagorean identity evaluations, standard acute angle calculations, and trigonometric ratios.

6 problems·20–25 min·★★★★☆
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  1. Problem 1
    Trigonometric Identities·★★★★☆
    Use Pythagorean identities to test $(\tan \theta + 6)(6\tan \theta + 1) = 37\tan \theta + \sec^2 \theta$.
    ▶Answer
    The statement is false.
    ▶Step-by-step solution
    1. The left side expands to $6\tan^2 \theta + 37\tan \theta + 6$.
    2. Using $\sec^2 \theta = 1 + \tan^2 \theta$, the right side is $\tan^2 \theta + 37\tan \theta + 1$.
    3. The two sides differ by $5(\tan^2 \theta + 1)$, so they are not identically equal.
    4. The statement is false.
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  2. Problem 2
    Special Angle Values·★★★☆☆
    Evaluate $(\sec 30^\circ + \tan 30^\circ) - (\operatorname{cosec} 60^\circ + \cot 60^\circ)$.
    ▶Answer
    The exact value is $0$.
    ▶Step-by-step solution
    1. Use the standard-angle values: $\sec 30^\circ = \frac{2\sqrt{3}}{3}, \tan 30^\circ = \frac{\sqrt{3}}{3}, \operatorname{cosec} 60^\circ = \frac{2\sqrt{3}}{3}, \cot 60^\circ = \frac{\sqrt{3}}{3}$.
    2. Substitute them into the expression: $(\sec 30^\circ + \tan 30^\circ) - (\operatorname{cosec} 60^\circ + \cot 60^\circ)$ = $(\frac{2\sqrt{3}}{3} + \frac{\sqrt{3}}{3}) - (\frac{2\sqrt{3}}{3} + \frac{\sqrt{3}}{3})$.
    3. Simplifying gives $0$.
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  3. Problem 3
    Trigonometric Ratios·★★★☆☆
    Given $\tan \theta = \sqrt{3}$, determine $\sin^2 \theta - \cos^2 \theta$.
    ▶Answer
    $\sin^2 \theta - \cos^2 \theta$ = $\frac{1}{2}$.
    ▶Step-by-step solution
    1. The condition identifies $\theta = 60^\circ$ for an acute angle.
    2. So $\sin^2 \theta - \cos^2 \theta$ = $\sin^2 60^\circ - \cos^2 60^\circ$.
    3. Using standard values gives $\frac{1}{2}$.
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  4. Problem 4
    Special Angle Values·★★★★☆
    Use standard-angle values to simplify $\frac{\cos 30^\circ}{\cos 60^\circ}$.
    ▶Answer
    The exact value is $\sqrt{3}$.
    ▶Step-by-step solution
    1. Use the standard-angle values: $\cos 30^\circ = \frac{\sqrt{3}}{2}, \cos 60^\circ = \frac{1}{2}$.
    2. Substitute them into the expression: $\frac{\cos 30^\circ}{\cos 60^\circ}$ = $\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}$.
    3. Simplifying gives $\sqrt{3}$.
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  5. Problem 5
    Trigonometric Identities·★★★★★
    Decide whether $\sqrt{(1 - \cos^2 \theta)\sec^2 \theta} = \tan \theta$ is true or false.
    ▶Answer
    The statement is true.
    ▶Step-by-step solution
    1. Use $1 - \cos^2 \theta = \sin^2 \theta$.
    2. Then the left side becomes $\sqrt{\sin^2 \theta\sec^2 \theta}$.
    3. Since the angle is acute, this equals $\sin \theta\sec \theta = \tan \theta$.
    4. So the statement is true.
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  6. Problem 6
    Trigonometric Ratios·★★★★☆
    Given $\cos A = \frac{15}{17}$, determine $\tan A$.
    ▶Answer
    $\tan A$ = $\frac{8}{15}$.
    ▶Step-by-step solution
    1. Since $\cos A = \frac{15}{17}$, take adjacent side = 15 and hypotenuse = 17.
    2. The opposite side is $\sqrt{17^2 - 15^2} = 8$.
    3. Therefore $\tan A$ = $\frac{8}{15}$ = $\frac{8}{15}$.
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