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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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