1.

`3sec^4theta+8=10sec^2theta` then find the value of `tantheta`

Answer» `3sec^4theta +8 = 10sec^2theta`
`=>3(1+tan^2theta)^2 +8 -10(1+tan^2theta) = 0`
Let `tan^2 theta = x`, then,
`=>3(1+x)^2+8 -10(1+x)= 0`
`=>3(1+x^2+2x)-10-10x +8 = 0`
`=>3x^2-4x +1 = 0`
`=>3x^2-3x -x +1 = 0`
`=>3x(x-1)-1(x-1) = 0`
`=>(3x-1)(x-1) = 0`
`=>x = 1/3 or x = 1`
`=>tan^2theta = 1/3 or tan theta = 1`
`=>tantheta = +-1/sqrt3 or tan theta = +-1.`


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