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प्रश्न
Prove that (sin θ + cosec θ)2 + (cos θ + sec θ)2 = 7 + tan2 θ + cot2 θ.
उत्तर १
L.H.S = (sin θ + cosec θ)2 + (cos θ + sec θ)2
= (sin2θ + cosec2θ + 2 sin θ cosec θ + cos2θ + sec2θ + 2cos θ sec θ)
= (sin2θ + cos2θ) + (cosec2θ + sec2θ) + 2 sin θ `(1/("sin"theta)) + 2 cos theta (1/("cos" theta))`
= (1) + (1 + cot2θ + 1 + tan2θ) + (2) + (2)
= 7 + tan2θ + cot2θ
= R.H.S
उत्तर २
L.H.S = (sin θ + cosec θ)2 + (cos θ + sec θ)2
= (sin2θ + cosec2θ + 2 sin θ cosec θ + cos2θ + sec2θ + 2cos θ sec θ)
= (sin2θ + cos2θ ) + 1 + cot2θ + 2 sin θ x `1/sin θ` + 1 + tan2 θ + 2cos θ. `1/cos θ`
= 1 + 1 + 1 + 2 + 2 + tan2 θ + cot2θ
= 7 + tan2 θ + cot2θ
= 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`
