InterviewSolution
Saved Bookmarks
| 1. |
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) 2A. `-1`B. 0C. 1D. 2 |
|
Answer» Correct Answer - 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` |
|