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प्रश्न
If sin θ + cos θ = `sqrt(3)`, then prove that tan θ + cot θ = 1.
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उत्तर
sin θ + cos θ = `sqrt(3)`
Squaring on both sides:
(sin θ + cos θ)2 = `(sqrt(3))^2`
sin2 θ + cos2 θ + 2 sin θ cos θ = 3
1 + 2 sin θ cos θ = 3
2 sin θ cos θ = 3 – 1
2 sin θ cos θ = 2
∴ sin θ cos θ = 1
L.H.S = tan θ + cot θ
= `sin theta/cos theta + cos theta/sin theta`
= `(sin^2 theta + cos^2 theta)/(sin theta cos theta)`
= `1/(sin theta cos theta)`
= `1/1` ...(sin θ cos θ = 1)
= 1 = R.H.S.
⇒ tan θ + cot θ = 1
L.H.S = R.H.S
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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`
