Question

Value of `(3+cot80^0cot20^0)/(cot80^0+cot20^0)` is equal to (a)`cot20^0` (b) `tan50^0` (c)`cot50^0` (d) `cotsqrt(20^0)`

Answer 1

`(3+cot80^@cot20^@)/(cot80^@+cot20^@)`
`=(3+(cos80^@/sin80^@ )(cos20^@/sin20^@))/((cos80^@/sin80^@)+(cos20^@/sin20^@))`
`=(3sin80^@sin20^@+cos80^@cos20^@)/(sin20^@cos80^@+sin80^@cos20^@)`
`=(2sin80^@sin20^@+cos(80^@-20^@))/(sin(20^@+80^@)`
`=(cos60^@ - cos100^@ +cos60^@)/(sin100^@)`
`=(2cos60^@ - cos100^@ )/(sin100^@)`
`=(2(1/2) - cos100^@ )/(sin100^@)`
`=(1 - cos100^@ )/(sin100^@)`
`=(2sin^2 50^@)/(2sin50^@cos50^@)`
`=sin50^@/cos50^@`
`= tan50^@`
So, option `(b)` is the correct option.