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If alpha+beta=pi/2a n dbeta+gamma=alpha,...

If `alpha+beta=pi/2a n dbeta+gamma=alpha,` then `tanalpha` equals `2(tanbeta+tangamma)` (b) `tanbeta+tangamma` `tanbeta+2tangamma` (d) `2tanbeta+tangamma`

A

`2(tanbeta+tangamma)`

B

`tanbeta+tangamma`

C

`tan beta+2 tangamma`

D

`2tanbeta+tangamma`

Text Solution

Verified by Experts

The correct Answer is:
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.`
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