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If sinx = 12/13, where 0°<x<90°, find the value of (1-cos²x)/(1-sin²30°)

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6. Given that cos⁡30∘=sin⁡60∘=32\cos 30^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2} and sin⁡30∘=cos⁡60∘=12\sin 30^\circ = \cos 60^\circ = \frac{1}{2}, evaluate (tan⁡60∘−1)(1−tan⁡30∘)\frac{(\tan 60^\circ - 1)}{(1 - \tan 30^\circ)}.

  1. Find the values of tan⁡60∘\tan 60^\circ and tan⁡30∘\tan 30^\circ.

    tan⁡60∘=sin⁡60∘cos⁡60∘=3212=3\tan 60^\circ = \frac{\sin 60^\circ}{\cos 60^\circ} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} tan⁡30∘=sin⁡30∘cos⁡30∘=1232=13\tan 30^\circ = \frac{\sin 30^\circ}{\cos 30^\circ} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}

  2. Substitute these values into the expression.

    (3−1)(1−13)\frac{(\sqrt{3} - 1)}{\left(1 - \frac{1}{\sqrt{3}}\right)}

  3. Simplify the denominator.

    1−13=33−13=3−131 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}}

  4. Substitute the simplified denominator back into the main expression and simplify.

    3−13−13=(3−1)×33−1=3\frac{\sqrt{3} - 1}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = (\sqrt{3} - 1) \times \frac{\sqrt{3}}{\sqrt{3} - 1} = \sqrt{3}

    The value is ∗3∗\boxed*{\sqrt{3}*}.

    7. If sin⁡x=1213\sin x = \frac{12}{13} where 0∘<x<90∘0^\circ < x < 90^\circ, find the value of (1−cos⁡2x)(1 - \cos^2 x).

  5. Use the trigonometric identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1.

    Rearranging the identity, we get 1−cos⁡2x=sin⁡2x1 - \cos^2 x = \sin^2 x.

  6. Substitute the given value of sin⁡x\sin x into the identity.

    1−cos⁡2x=sin⁡2x=(1213)21 - \cos^2 x = \sin^2 x = \left(\frac{12}{13}\right)^2

  7. Calculate the square.

    (1213)2=122132=144169\left(\frac{12}{13}\right)^2 = \frac{12^2}{13^2} = \frac{144}{169}

    The value is ∗144169∗\boxed*{\frac{144}{169}*}.

    8. Without using tables or calculators, calculate the value of (1+cos⁡230∘)(1−sin⁡230∘)\frac{(1 + \cos^2 30^\circ)}{(1 - \sin^2 30^\circ)}.

  8. Recall the values of cos⁡30∘\cos 30^\circ and sin⁡30∘\sin 30^\circ.

    cos⁡30∘=32\cos 30^\circ = \frac{\sqrt{3}}{2} sin⁡30∘=12\sin 30^\circ = \frac{1}{2}

  9. Substitute these values into the expression.

    (1+(32)2)(1−(12)2)\frac{\left(1 + \left(\frac{\sqrt{3}}{2}\right)^2\right)}{\left(1 - \left(\frac{1}{2}\right)^2\right)}

  10. Calculate the squares.

    (32)2=34\left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} (12)2=14\left(\frac{1}{2}\right)^2 = \frac{1}{4}

  11. Substitute the squared values back into the expression and simplify the numerator and denominator.

    (1+34)(1−14)=(44+34)(44−14)=7434\frac{\left(1 + \frac{3}{4}\right)}{\left(1 - \frac{1}{4}\right)} = \frac{\left(\frac{4}{4} + \frac{3}{4}\right)}{\left(\frac{4}{4} - \frac{1}{4}\right)} = \frac{\frac{7}{4}}{\frac{3}{4}}

  12. Divide the numerator by the denominator.

    7434=74×43=73\frac{\frac{7}{4}}{\frac{3}{4}} = \frac{7}{4} \times \frac{4}{3} = \frac{7}{3}

    The value is ∗73∗\boxed*{\frac{7}{3}*}.

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