Question

Express the trignometric ratios sinA, secA nd tanA in terms of cotA​

Answer 1

Explanation:

HOPE YOU UNDERSTAND THIS

Answer 2

Answer:

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sinA = P/H = BC/AC

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Let BC = x and AB = xcotA

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Now in Δ ABC, by pythagorus theorem

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P² + B² = H²

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$\longrightarrow$ x² + (xcotA)² = H²

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= √(x² + x²cot²A) = H

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$\longrightarrow$ √{x²(1 + cot²A)} = H

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$\longrightarrow$ x√(1 + cot²A) = H

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$\longrightarrow$ sinA = P/H = x/x(1 + cos²A) = 1/(1 + cos²A)

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$\longrightarrow$ secA = H/B = x(1 + cot²A)/xcotA = (1 + cot²A)/cotA

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$\longrightarrow$ tanA = P/B = x/xcotA = 1/cotA