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प्रश्न
Prove that `sqrt((1 + cos A)/(1 - cos A)) = "cosec" A + cot A`.
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उत्तर
L.H.S. = `sqrt((1 + cos A)/(1 - cos A))`
= `sqrt((1 + cos A)/(1 - cos A) xx (1 + cos A)/(1 + cos A))` ...[On rationalising the denominator]
= `sqrt((1 + cos A)^2/(1 - cos^2 A))`
= `sqrt((1 + cos A)^2/(sin^2 A)` ...`[(∵ sin^2A + cos^2A = 1),(∴ 1 - cos^2A = sin^2A)]`
= `(1 + cos A)/(sin A)`
= `1/(sin A) + (cos A)/(sin A)`
= cosec A + cot A
= R.H.S.
∴ `sqrt((1 + cos A)/(1 - cos A)) = "cosec" A + cot A`
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Solution:
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`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
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But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
