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The value of cos^2 10^0-cos10^0cos50^0+c...

The value of `cos^2 10^0-cos10^0cos50^0+cos^2 50^0` is equal to `4/3` (b) `1/3` (c) `3/4` (d) `3`

A

`(3)/(2) (1+cos20^(@))`

B

`(3)/(4)+cos20^(@)`

C

`3//2`

D

`3//4`

Text Solution

Verified by Experts

The correct Answer is:
D
We have, `cos^(2)10^(@)-cos10^(@)cos50^(@)+cos^(2)50^(@)`
`=(1)/(2)[2cos^(2)10^(@)-2cos10^(@)cos50^(@)+2cos^(2)50^(@)]`
`=(1)/(2)[1+cos20^(@)-(cos60^(@)+cos40^(@))+1+cos 100^(@)] " "[because 2 cos^(2) A = 1 + cos 2 A and 2 cos A and B = cos(A + B) + cos (A - B)]`
`=(1)/(2)[2+cos20^(@)+cos100^(@)-(1)/(2)-cos40^(@)][because cos 60^(@)=(1)/(2)]`
`=(1)/(2)[2+cos20^(@)+cos100^(@)-(1)/(2)-cos40^(@)][because cos 60^(@)=(1)/(2)]`
`=(1)/(2)[(3)/(2)-2sin.(20^(@)+40^(@))/(2)sin.(20^(@)-40^(@))/(2)+cos100^(@)]`
`[because cos C - cos D = -2sin.(C+D)/(2)sin.(C-D)/(2)]`
`=(1)/(2)[(3)/(2)-2sin30^(@)sin(-10^(@))+cos(90^(@)+10^(@))]`
`=(1)/(2)[(3)/(2)+sin10^(@)-sin10^(@)][because cos (90^(@)+theta)=-sintheta]`
`=(1)/(2)xx(3)/(2)=(3)/(4)`
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