( sin theta + sec theta ) whole square + ( cos theta + cosec theta) whole square = ( 1+ sec theta . cosec theta ) whole suare
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Answered by
907
Let angle theta be represented by A.
(sinA + Sec A)² + (cos A + cosec A)²
= [ (SinA CosA + 1)²/CosA ]² + [ (CosA SinA + 1)² / Sin²A ]
= (1+ SinA CosA)² * [1 / Cos²A + 1/sin²A ]
= (1 + SinA CosA)² * [ Cos²A + Sin²A]/ [cos²A * Sin²A ]
= ( 1 + SinA CosA)²/ (cosA * SinA)²
= [ 1/cosA * 1/sinA + 1 ] ²
= [ Sec A Cosec A + 1 ]²
(sinA + Sec A)² + (cos A + cosec A)²
= [ (SinA CosA + 1)²/CosA ]² + [ (CosA SinA + 1)² / Sin²A ]
= (1+ SinA CosA)² * [1 / Cos²A + 1/sin²A ]
= (1 + SinA CosA)² * [ Cos²A + Sin²A]/ [cos²A * Sin²A ]
= ( 1 + SinA CosA)²/ (cosA * SinA)²
= [ 1/cosA * 1/sinA + 1 ] ²
= [ Sec A Cosec A + 1 ]²
Answered by
762
See the simplified step wise answer in the attachment
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