Question 90

If Cot A = $$\frac{12}{5}$$ then $$(Sin A + Cos A) \times Cosec$$ A is?

Solution

Cot A = $$\frac{12}{5}$$

$$\frac{B}{P}=\frac{12}{5}$$

$$H^2=12^2+5^2=169$$

H = 13

$$(Sin A + Cos A) \times Cosec$$ A

$$\left(\frac{P}{H}+\frac{B}{H}\right)\times\ \frac{H}{P}$$

$$\left(\frac{5}{13}+\frac{12}{13}\right)\times\ \frac{5}{13}=\frac{17}{13}\times\ \frac{5}{13}=\frac{17}{5}$$

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