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[Cosecα – sinα]= a 3 ;[Secα -cosα]= b 3 a 2 b 2 [ a 2 +b 2 ]= ? show the working

[Cosecα – sinα]= a3  ;[Secα -cosα]= b3


a2b2[ a2+b2]= ?


show the working 

Grade:

1 Answers

sanatan sharma
26 Points
9 years ago

cosec θ - sin θ =a3                                                                            (1)

sec θ - cos θ = b3                                                                    (2)

from (1)         1/sin θ - sin θ = a3    or   (1- sin2θ)/sin θ = a3           or      

cos2θ/sin θ = a3                                                                           (3)

similarly from (2)     we get               sin2θ/cos θ = b3                          (4)

from (4)   cos θ = sin2θ/b3

substituting in  (3)   we get

sin4θ/b6 = a3sin θ 

therefore sin θ = ab                                                                (5)

cos θ = sin2θ/b3              so               cos θ = a2b                         (6)

squaring and adding (5) and (6)

sin2θ +cos2θ = a2b4 +a4b2

1 = a2b4 +a4b2



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