Question

If sin A+cosA=m
seca+cosec An
Prove that n(m2_1)
= 2m

Answer 1

Answer:

LHS : n(m2 - 1)

= secA +cosecA [ (sinA +cosA)2 - 1]

= secA + cosecA [ sin2A + cos2A + 2 sinA .cosA -1]

= secA +cosecA [ 1 +2 sinA . cosA -1] [sin2A +cos2A =1]

= secA +cosecA [ 2sinA cosA]

= 2 secA sinA cosA + 2 cosecA sinA cosA

= 2sinA + 2 cosA [secA*cosA=1 and cosecA*sinA=1]

= 2(sinA +cosA)

= 2 m [sinA +cosA =m]

= RHS

Answer 2

Answer:

n[m^2-1]=2m

Step-by-step explanation:

sec A+cosecA [(sinA+cosA)^2 - 1]

= 1/cosA +1/sinA[sin^2A +cos^2A+2sinAcosA-1]

=sinA+cosA/sinAcosA [1+2sinAcosA-1]

sinA+cosA/sinAcosA[2 sinAcosA]

= 2[sinAcosA]

hence proved

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