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(cosx)/cos(x-2y)=lambda=>tan(x-y)tany=...

`(cosx)/cos(x-2y)=lambda=>tan(x-y)tany=`

A

`(1+lambda)/(1-lambda)`

B

`(1-lambda)/(1+lambda)`

C

`(lambda)/(1+lambda)`

D

`(lambda)/(1-lambda)`

Text Solution

Verified by Experts

The correct Answer is:
B
`tan(x-y)tany=(sin(x-y)siny)/(cos(x-y)cosy)xx(2)/(2)`
`=(cos(x-2y)-cos(x))/(cos(x-2y)+cos(x))`
`=(1-(cosx)/(cos(x-2y)))/(1+(cosx)/(cos(x-2y)))`
`=(1-lambda)/(1+lambda)" "[because lambda=(cosx)/(cos(x-2y))]`
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MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS-FACTORIZATION FORMULAE -EXERCISE 2
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