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Neokaal Study

Introduction to Trigonometry

Competency Practice

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

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  1. Use Pythagorean identities to test $(\tan \theta + 6)(6\tan \theta + 1) = 37\tan \theta + \sec^2 \theta$.

  2. Evaluate $(\sec 30^\circ + \tan 30^\circ) - (\operatorname{cosec} 60^\circ + \cot 60^\circ)$.

  3. Given $\tan \theta = \sqrt{3}$, determine $\sin^2 \theta - \cos^2 \theta$.

  4. Use standard-angle values to simplify $\frac{\cos 30^\circ}{\cos 60^\circ}$.

  5. Decide whether $\sqrt{(1 - \cos^2 \theta)\sec^2 \theta} = \tan \theta$ is true or false.

  6. Given $\cos A = \frac{15}{17}$, determine $\tan A$.

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Answer Key

  1. The statement is false.

    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.

  2. The exact value is $0$.

    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 $0$.

  3. $\sin^2 \theta - \cos^2 \theta$ = $\frac{1}{2}$.

    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 $\frac{1}{2}$.

  4. The exact value is $\sqrt{3}$.

    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 $\sqrt{3}$.

  5. The statement is true.

    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.

  6. $\tan A$ = $\frac{8}{15}$.

    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 $\tan A$ = $\frac{8}{15}$ = $\frac{8}{15}$.