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If tanalpha=(1)/(7) and tanbeta=(1)/(3),...

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

A

`sin2beta`

B

`sin4beta`

C

`sin3beta`

D

Note of these

Text Solution

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The correct Answer is:
B
`cos2alpha=(1-t^(2))/(1+t^(2))=(24)/(25) " " ["here",t=tanalpha]`
`sin2beta=(2T)/(1+T^(2))=(3)/(5)`
`rArrcos2beta=(4)/(5) " " [Ttanbeta]`
`thereforesin4beta=2sin2betacos2beta`
`=2*(3)/(5)*(4)/(5)=(24)/(25)`
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