Given that sin⁡A=p\sin A = p, where A is obtuse, find

  1. (i)

    cos⁡A\cos A.

  2. (ii)

    cot⁡A\cot A.

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Worked solution

1(i)

1. Use the Pythagorean identity to express $\cos^2 A$ in terms of $p$

sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1
cos⁡2A=1−sin⁡2A\cos^2 A = 1 - \sin^2 A
cos⁡2A=1−p2\cos^2 A = 1 - p^2

2. Determine the sign of $\cos A$ using the given quadrant

Since AA is obtuse (90∘<A<180∘90^\circ < A < 180^\circ), cos⁡A\cos A must be negative.
cos⁡A=−1−p2\cos A = -\sqrt{1 - p^2}

Answer

−1−p2-\sqrt{1 - p^2}

1(ii)

1. Express $\cot A$ in terms of $\sin A$ and $\cos A$

cot⁡A=cos⁡Asin⁡A\cot A = \frac{\cos A}{\sin A}

2. Substitute the values of $\sin A$ and $\cos A$ to find $\cot A$

From part (i), cos⁡A=−1−p2\cos A = -\sqrt{1 - p^2}.
cot⁡A=−1−p2p\cot A = \frac{-\sqrt{1 - p^2}}{p}

Answer

−1−p2p-\frac{\sqrt{1 - p^2}}{p}

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