please evaluate

Asked by sunil_0010 | 31st Oct, 2009, 08:50: PM

Expert Answer:

LHS =

(tanA-tanB)2+(1+tanA.tanB)2=

(sinA/cosA - sinB/cosB)2 + (1 + sinAsinB/(cosAcosB))2 =

[(sinAcosB - sinBcosA)/(cosAcosB)]2 + [(cosAcosB + sinAsinB)/(cosAcosB)]2 =

[sin(A - B)/(cosAcosB)]2 + [cos (A - B)/(cosAcosB)]2 =

[sin2(A - B) + cos2 (A - B)]/(cosAcosB)2 =

1/(cosAcosB)2 = sec2Asec2B

= RHS

Regards,

Team,

TopperLearning.

 

 

 

Answered by  | 31st Oct, 2009, 10:22: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.