1.

Prove that `1+cot theta

Answer» `1+cot theta - cot(theta/2) = 1+(cot^2(theta/2) - 1)/(2cot(theta/2)) - cot(theta/2)`
`=(2cot(theta/2)+cot^2(theta/2)-1 -2cot^2(theta/2))/(2cot(theta/2))`
`=-(cot^2(theta/2)-2cot(theta/2)+1)^2/(2cot(theta/2))`
`=-(cot theta-1)^2/(2cot(theta/2))`
As `-(cot (theta/2)-1)^2/(2cot(theta/2))` is always less than or equal to `0`,
`:. 1+cot theta - cot(theta/2) le 0`
`=>1+cot theta le cot(theta/2).`
Now, when equality sign holds,
`(cot(theta/2) - 1)^2 = 0`
`=>cot (theta/2) = 1`
`=>theta/2 = pi/4`
`=>theta = pi/2.`


Discussion

No Comment Found

Related InterviewSolutions