RRB NTPC 30 March 2016 Shift 1

Question 66

If $$\cos \theta = \frac{4}{5}$$, then $$(\sec \theta + \cosec \theta)$$ =

Solution

$$\cos\ \theta\ =\frac{B}{H}=\frac{4}{5}$$

P= $$\sqrt{\ 5^2-4^2}=\sqrt{\ 9}=3$$

$$(\sec \theta + \cosec \theta)$$

$$\frac{H}{B}+\frac{H}{P}$$

$$\frac{5}{4}+\frac{5}{3}=\frac{35}{12}$$.

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