Prove: sec a× under root 1-sin square a =1
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secA × √{ 1 - sin²A}
We know, 1 - sin²A = cos²A
secA × √cos²A
secA × cosA
We know, secA = 1/cosA
1/cosA × cosA
1
Hence, proved.
We know, 1 - sin²A = cos²A
secA × √cos²A
secA × cosA
We know, secA = 1/cosA
1/cosA × cosA
1
Hence, proved.
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