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It cos(alpha+beta)=4/5,sin(alpha-beta)=5...

It `cos(alpha+beta)=4/5,sin(alpha-beta)=5/(13)a n dalpha,beta` lie between `0a n dpi/4` , prove that `tan2alpha=(56)/(33)`

A

`(63)/(52)`

B

`(63)/(16)`

C

`(21)/(16)`

D

`(33)/(52)`

Text Solution

Verified by Experts

The correct Answer is:
B
Given, `sin(alpha - beta) = (5)/(13)`
and `cos(alpha + beta) = (3)/(5)`, where `alpha, beta in (0,(pi)/(4))`
Since, `0 lt alpha lt (pi)/(4)` and `0 lt beta lt (pi)/(4)`
`therefore 0 lt alpha + beta lt (pi)/(4) + (pi)/(4) = (pi)/(2)`
`implies 0 lt alpha + beta lt (pi)/(2)`
Also, `-(pi)/(4) lt -beta lt 0`
`therefore 0 - (pi)/(4) lt alpha - beta lt (pi)/(4) + 0`
`implies -(pi)/(4) lt alpha - beta lt (pi)/(4)`
`therefore alpha + beta in (0, (pi)/(2)) and alpha - beta in (-(pi)/(4), (pi)/(4))`
But `sin(alpha - beta) gt 0`, therefore `alpha - beta in (0, (pi)/(4))`
Now, `sin(alpha-beta) = (5)/(13)`
`implies tan(alpha - beta)=(5)/(12)" (i)"`
and `cos(alpha + beta) = (3)/(5)`
`implies tan (alpha + beta) = (4)/(3) " (ii)"`
Now, `tan(2alpha) = tan [(alpha + beta) + (alpha-beta)]`
`=(tan(alpha+beta)+tan(alpha-beta))/(1-tan(alpha+beta)tan(alpha-beta))=((4)/(3)+(5)/(12))/(1-(4)/(3)xx(5)/(12))" [from Eqs. (i) and (ii)]"`
`=(48 + 15)/(36-20)=(63)/(16)`
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