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If ( cos x - cos alpha)/(cos x - cos b...

If ` ( cos x - cos alpha)/(cos x - cos beta) = ( sin^2 alpha cos beta)/(sin^2 beta cos alpha)` then cos x =

A

`cos x=(cos alpha+cos beta)/(1-cos alpha cos beta)`

B

`cos x=(cos alpha+cos beta)/(1-cos alpha cos beta)`

C

`tan.(x)/(2)=tan.(alpha)/(2)tan.(beta)/(2)`

D

`tan.(x)/(2)=-tan.(alpha)/(2)tan.(beta)/(2)`

Text Solution

Verified by Experts

The correct Answer is:
A, C, D
From the given redults
`cos x sin^(2)bet cos alpha - cos^(2) alpha sin^(2)beta`
`= sin^(2) alpha cos beta cos x - sin^(2) alpha cos^(2) beta`
`rArr os x[cos alpha sin^(2)beta - sin^(2) alpha cos beta]`
`= cos^(2)alpha sin^(2)beta-sin^(2)alpha cos^(2) beta`
`rArr cos x=(cos^(2)alpha[1-cos^(2)beta]-(1-cos^(2)alpha)cos^(2)beta)/(cos alpha[1-cos^(2)beta]-[1-cos^(2)alpha[cos beta)`
`=(cos^(2)alpha-cos^(2)beta)/((cos alpha-cos beta)(cos alpha cos beta+1))`
`=(cos alpha + cos beta)/(1+cos alpha cos beta)`
`rArr (co x)/(1)=(cos alpha + cos betA)/(1+cos alpha cos beta)`
`rarr (1-cos x)/(1+cos x)=(1+cos alpha cos beta - cos alpha - cos beta)/(1+cos lapha cos beta + cos alpha + cos beta)`
`=((1-cos alpha)(1-cos beta))/((1+cos alpha)(1+cos beta))`
`rArr "tan"^(2)(x)/(2)="tan"^(2)(alpha)/(2)."tan"^(2)(beta)/(2)`
`rArr tan.(x)/(2)= pm tan.(alpha)/(2)tan.(beta)/(2)`
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