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प्रश्न
Prove the following trigonometric identities.
`1/(1 + sin A) + 1/(1 - sin A) = 2sec^2 A`
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उत्तर
We have to prove `1/(1 + sin A) + 1/(1 - sin A) = 2sec^2 A`
We know that, `sin^2 A + cos^2 A = 1`
So,
`1/(1 + sin A) + 1/(1 - sin A) =((1 - sin A) + (1 + sin A))/((1 + sin A)(1 - sin A))`
`= (1 - sin A + 1+ sin A)/(1 - sin^2 A)`
`= 2/cos^2 A`
`= 2 sec^2 A`
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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`
