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If cosx-sinalphacotbetasinx=cosa , then ...

If `cosx-sinalphacotbetasinx=cosa ,` then the value of `tan(x/2)` is `-tan(alpha/2)cot(beta/2)` (b) `tan(alpha/2)tan(beta/2)` `-cot((alphabeta)/2)tan(beta/2)` (d) `cot(alpha/2)cot(beta/2)`

A

`-tan (alpha2)cot(beta2)`

B

`tan (alpha//2)tan(beta//2)`

C

`-cot(alpha//2)tan(beta//2)`

D

`cot(alpha//2)cot(beta//2)`

Text Solution

Verified by Experts

The correct Answer is:
A, B

`cos x - sin alpha cos beta sinx = cos alpha `
`rArr (1-tan^(2)(x//2))/(1+tan^(2)(x//2)) - sin alpha cos beta (2tan(x//2))/(1+tan^(2)(x//2)) = cos alpha`
`rArr tan^(2) ""(x)/(2) (1+ cosalpha ) + 2 sin alpha cos beta tan ""(x)/(2) - (1- cos alpha) =0`
`rArr tan ^(2)""(x)/(2) + (2sinalpha cos beta)/(1+ cos alpha) tan""(x)/(2) - (1-cos alpha)/( 1+cos alpha) =0`
`rArr tan^(2)""(x)/(2) + 2 tan""(alpha)/(2) cos beta tan ""(x)/(2) - tan^(2) ""(alpha)/(2)=0`
`rArr tan^(2""(x)/(2) + 2 tan""(alpha)/(2)*(1)/(2)(cot""( beta)/(2) - tan ""(beta)/(2)) tan ""(x)/(2) - tan^(2) (alpha)/(2) =0`
`rArr (tan ""(x)/(2) + cot""(beta)/(2) tan""(alpha)/(2)) (tan""(x)/(2) - tan""(beta)/(2) tan""(alpha)/(2)) =0`
`rArr tan((x)/(2)) = - tan((alpha)/(2)) cot((beta)/(2))`
or `" " tan ((x)/(2)) = tan((alpha)/(2) tan((beta)/(2))`
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