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Correct Answer – C
We have,
`beta+gamma+alpha`
`impliesgamma=alpha=beta`
`tangamma=tan(alpha-beta)`
`impliestangamma=(tanalpha-tanbeta)/(1+tanalphatanbeta)`
`impliestangamma=(tanalpha-tanbeta)/(1+tanalphacot alpha)” “[becausealpha+beta=(pi)/(2)becausebeta=(pi)/(2)-alpha]`
`impliestangamma=1/2(tangamma-tanbeta)impliestanalpha=tanbeta+2tangamma.`
Correct Answer – c
Given, `alpha+beta=pi//2`
`implies alpha=(pi//2)-beta`
`implies tan alpha = tan (pi//2-beta)`
`implies tan alpha = cot beta`
`implies tan alpha tan beta = 1`
Again, `beta + gamma = alpha ” [given]”`
`implies gamma = (alpha – beta)`
`implies tan gamma = tan(alpha – beta)`
`implies tan gamma = (tanalpha-tanbeta)/(1+tanalphatanbeta)`
`implies tan gamma = (tan alpha-tanbeta)/(1+1)`
`therefore 2 tan gamma = tan alpha – tan beta`
`implies tan alpha=tanbeta + 2 tan gamma`