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If sin A = x, then the value of tan A sec A1). \(\frac{x}{{\sqrt {1 + {x^2}} }}\)2). \(\sqrt {1 - {x^2}} \)3). \(\sqrt {\frac{{1 - {x^2}}}{{{x^2}}}}\)4). \(\frac{x}{{1 - {x^2}}}\) |
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Answer» sin A = Perpendicular/Hypotenuse = X/1 From PYTHAGORAS theorem ⇒ Base = √(1 – x2) sec A = Perpendicular/Base = $(\frac{1}{{\SQRT {1 - {x^2}} }})$ tan A = Hypotenuse/Base = $(\frac{x}{{\sqrt {1 - {x^2}} }})$ tan A sec A = $(\frac{x}{{1 - {x^2}}})$ |
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