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CBSE Class 10 Answered

please evaluate
Asked by sunil_0010 | 31 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

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Team,

TopperLearning.

 

 

 

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