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If alpha,beta,gamma are ccute angles and...

If `alpha,beta,gamma` are ccute angles and `costheta=sinbeta//sinalpha,cosvarphi=singammasinalphaa n dcos(theta-varphi)=sinbetasingamma` , then the value of `tan^2alpha-tan^2beta-tan^2gamma` is equal to `-1` (b) `0` (c) `1` (d) 2

A

`-1`

B

0

C

1

D

2

Text Solution

Verified by Experts

The correct Answer is:
B
From the third relation, we get
`cos theta cos phi + sin theta sin phi = sin beta sin gamma`
`rArr sin^(2)theta sin^(2)phi=(cos theta cos phi - sin beta sin gamma)^(2)`
`rArr (1-(sin^(2)beta)/(sin^(2)alpha))(1-(sin^(2)gamma)/(sin^(2)alpha))=((sin beta sin gamma)/(sin^(2)alpha)-sin beta sin gamma)^(2)`
`rArr (sin^(2)alpha-sin^(2)beta)(sin^(2)alpha-sin^(2)gamma)=sin^(2)beta sin^(2)gamma(1-sin^(2)alpha)^(2)`
`rArr sin^(4)alpha(1-sin^(2)beta sin^(2)sin^(2)gamma)-sin^(2)alpha(sin^(2)beta + sin^(2)gamma-2 sin^(2)beta sin^(2)gamma)=0`
`therefore sin^(2)alpha=(sin^(2)beta + sin^(2)gamma -2sin^(2)beta sin^(2)gamma)/(1-sin^(2)beta si8n^(2)gamma)`
and `cos^(2)alpha=(1-sin^(2)beta-sin^(2)gamma + sin^(2)beta sin^(2)gamma)/(1-sin^(2)beta sin^(2)gamma)`
`rArr tan^(2)alpha=(sin^(2)beta -sin^(2)beta sin^(2)gamma+sin^(2)gamma-sin^(2)beta sin^(2)gamma)/(cos^(2)beta - sin^(2)gamma(1-sin^(2)beta))`
`=(sin^(2)beta cos^(2)gamma + cos^(2)beta sin^(2)gamma)/(cos^(2)beta cos^(2)gamma)`
`= tan^(2) bet + tan^(2) gamma`
`rArr tan^(2) alpha - tan^(2) beta ` - `tan^(2) gamma = 0`
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