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Find the value of other five trigonometric function `sinx=3/5`, x lies in second quadrant. |
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Answer» Correct Answer - c Given that `sinx=3/5` and x lies in second quadrant. `sinx=3/5` `therefore "cosec"x = 1/(sinx) = 1/(3//5)=1 xx 5/3=5/3` `therefore "cosec"x=5/3`. `(therefore` x lies in second quadrant, so sin ratio will be positive and the remaining ratio tan, cot, cos and sec will be negative). `therefore sin^(2)x+cos^(2)=1 rArr (3/5)^(2) + cos^(2)x=1` `rArr 9/25 + cos^(2)x=1` `cos^(2)x=1-9/25 rArr cos^(2)x=16/25` `rArr cosx =+-4/5` `therefore secx=1/(cosx)` `rArr tanx=(sinx)/(cosx) = (3//5)/(-4//5) = 3/5 xx (-5/4) = -3/4` `therefore cotx=(cosx)/(sinx)` `therefore cotx = (-4//5)/(3//5) = -4/5 xx 5/3 =-4/3` Therefore, `"cosec"x=5/3, cosx=-4/5`, `secx=-5/4, tanx=-3/4` and `cotx=-4/3` Ans. |
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