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प्रश्न
Prove the following trigonometric identities
`((1 + sin theta)^2 + (1 + sin theta)^2)/(2cos^2 theta) = (1 + sin^2 theta)/(1 - sin^2 theta)`
उत्तर
LHS = `(1 sin^2 theta + 2 sin theta + 1 + sin^2 theta - 2 sin theta)/(2 cos theta)`
`=> (2(1 + sin^2 theta))/(2 cos^2 theta) => (1 + sin^2 theta)/(1 - sin^2 theta)` `[∵ cos^2 theta = 1 - sin^2 theta]`
∴ LHS = RHS Hence proved
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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`
