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प्रश्न
Prove the following identity :
`1/(sinA + cosA) + 1/(sinA - cosA) = (2sinA)/(1 - 2cos^2A)`
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उत्तर
LHS = `1/(sinA + cosA) + 1/(sinA - cosA)`
= `(sinA - cosA + sinA + cosA)/(sin^2A - cos^2A)`
= `(2sinA)/(1 - cos^2A - cos^2A) = (2sinA)/(1 - 2cos^2A)`
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संबंधित प्रश्न
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sec2 A . cosec2 A = tan2 A + cot2 A + 2
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If tan θ = `13/12`, then cot θ = ?
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Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
