Question 87

If √$$(sec^2A-1)/secA$$ = x, then the value of x is

Solution

Expression : $$\frac{\sqrt{sec^2A-1}}{secA}$$

$$\because$$ $$(sec^2A-tan^2A=1)$$

= $$\frac{\sqrt{tan^2A}}{secA} = \frac{tanA}{secA}$$

= $$(\frac{sinA}{cosA})\div(\frac{1}{cosA})$$

= $$(\frac{sinA}{cosA})\times(cosA) = sinA$$

=> Ans - (C)

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