# Prove that: (SinA+secA)square+(cosA+cosecA)square=(1+secA×cosecA)square

1
by char

2015-04-25T08:45:14+05:30

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(sin A + sec A)² + (Cos A + Cosec A)²
= Sin² A + sec² A + 2 Sin A Sec A + Cos² A + Cosec² A + 2 Cos A Cosec A
= Sin²A + COs²A + Sec²A + Cosec²A + 2 Sin A SecA + 2 Cos A Cosec A
= 1  +  [ 1/Cos²A + 1/ Sin²A ] +  [ 2 Sin A / Cos A + 2 Cos A / SIn A ]
= 1 + (Sin²A + COs²A)/ [Cos²A Sin²A ]  + 2 [ SIn² A + Cos²A ] / [ SinA CosA
= 1 + 1/Cos²A 1/Sin²A  + 2 1/SinA  1/CosA
= 1 + Sec²A Cosec²A + 2 COsecA Sec A
= (1 + SecA CosecA )²

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alternately,

(sin A + sec A)² + (Cos A + Cosec A)²
=  (Sin A + 1/Cos A)² + (COs A + 1/ SinA)²
=   (Sin A Cos A  + 1)² / Cos² A    +  (SinA COsA + 1)² / Sin² A
= [ SIn A Cos A + 1]² [ 1/Cos² A + 1/Sin² A ]
= [  SIn A Cos A + 1]² [ Cos² A + Sin² A ] / [Sin²A Cos² A ]
= [ Sin A Cos A + 1 ]²  /  [Sin²A Cos² A ]
= [ (Sin A Cos A + 1) / (Sin A Cos A )]²
=  (1 + 1/Sin A  1/Cos A)²
= (1 + Sec A Cosec A)²