If `tanalpha =(1)/(7) and tanbeta =(1)/(3) , then, cos2alpha` is equal to

Options

A. `sin2beta`
B. `sin4beta`
C. `sin2beta`
D. `cos2beta`

Answer

B

Explanation

Given that, `tanalpha=(1)/(7) and tanbeta=(1)/(3)` ltBrgt `" "cos2alpha=(1-(1)/(49))/(1+(1)/(49))=((48)/49)/((50)/(49))`
`" "=(48)/(50)=(24)/(25)`
`rArr" "cos2alpha=(24)/(25)" "...(i)`
We know that, `" "sin4beta=(2tan2beta)/(1+tan^(2)2beta)" "…(ii)`
and `" "tan2beta=(2tanbeta)/(1-tan^(2)beta)=(2xx(1)/(3))/(1-(1)/(9))`
`" "=((2)/(3))/((8)/(9))=(2xx9)/(3xx8)=(3)/(4)`
From Eq, (ii),
`" "sin4beta=(2xx(3)/(4))/(1+(9)/(16))=((6)/(4))/((25)/(16))=(6xx16)/(4xx25)`
`rArr" "sin4beta=(24)/(25)`
`rArr" "sin4beta=cos2alpha`
`therefore" "cos2alpha=sin4beta" "` [from Eq. (i)]