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CBSE Class 10 Answered

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Asked by | 06 Jun, 2008, 07:25: PM
answered-by-expert Expert Answer

(cos3A+sin3A )/( cosA+ sinA)+(cos3A-sin3A) / (cosA-sinA)

= [(cosA+sinA)(cos2A+sin2A-cosAsinA)] / ( cosA+ sinA)   +  [(cosA-sinA)(cos2A+sin2A+cosAsinA)] / ( cosA- sinA)

= (cos2A+sin2A-cosAsinA) + (cos2A+sin2A+cosAsinA)]

= cos2A+sin2A-cosAsinA + cos2A+sin2A+cosAsinA

= 1 -cosAsinA + 1 + cosAsinA

= 2

Answered by | 07 Jun, 2008, 08:09: PM

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