First slide

Trigonometric Identities

Question

$\cos e c A = 4 p + \frac{1}{16 p} \Rightarrow \cos e c A + \cot A =$

Moderate

A

8p
B

$\frac{1}{8 p}$
C

$- 8 p (o r) \frac{1}{8 p}$
D

$8 p (o r) \frac{1}{8 p}$

Solution

Use $(\cos e c^{2} A - \cot^{2} A) = 1$
If $\cos e c A = (4 P + \frac{1}{16 P})$ then $cotA = \pm (4 P - \frac{1}{16 P})$

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