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प्रश्न
Show that `(cos^2(45^circ + θ) + cos^2(45^circ - θ))/(tan(60^circ + θ) tan(30^circ - θ)) = 1`
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उत्तर
L.H.S = `(cos^2(45^circ + θ) + cos^2(45^circ - θ))/(tan(60^circ + θ) xx tan(30^circ - θ))`
= `(cos^2(45^circ + θ) + [sin{90^circ - (45^circ - θ)}]^2)/(tan(60^circ + θ) xx cot{90^circ - (30^circ - θ)})` ...[∵ sin (90° – θ) = cos θ and cot (90° – θ) = tan θ]
= `(cos^2(45^circ + θ) + sin^2(45^circ + θ))/(tan(60^circ + θ) xx cot(60^circ + θ))` ...[∵ sin2θ + cos2θ = 1]
= `1/(tan(60^circ + θ)) xx 1/(tan(60^circ + θ))` ...`[∵ cot θ = 1/tan θ]`
= 1
= R.H.S
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Activity:
L.H.S = `square`
= `cos^2theta xx square .....[1 + tan^2theta = square]`
= `(cos theta xx square)^2`
= 12
= 1
= R.H.S
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Find the value of sin2θ + cos2θ
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`
