prove that sintheta + sectheta the whole square + costheta + cosectheta the whole square=1 + sectheta cosectheta the whole square
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Answer:
Step-by-step explanation:
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
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