Find the value of other five trigonometric function `sinx=3/5`, x lies in second quadrant.

Answer

c

Explanation

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.