Question
Solution
Given: Triangle ABC is right-angled at B So, ∠A + ∠C = 90° ⇒ sin A = cos C and cos A = sin C Given: cosec A = 2√2 ⇒ sin A = 1 / (2√2) = √2 / 4 Then:  cos A = √(1 − sin²A) = √(1 − (2/16)) = √(14/16) = √14 / 4  sin C = cos A = √14 / 4  cos C = sin A = √2 / 4 Now compute: sin A · cos C + cos A · sin C = (√2/4)(√2/4) + (√14/4)(√14/4) = 2/16 + 14/16 = 16/16 = 1
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