prove the...following....look in the image
Answers
Answer:hope that helps...please mark as brainliest
Step-by-step explanation:
(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
Hope it helps.............
