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प्रश्न
Show that `(cos^2(45^circ + theta) + cos^2(45^circ - theta))/(tan(60^circ + theta) tan(30^circ - theta))` = 1
उत्तर
L.H.S = `(cos^2(45^circ + theta) + cos^2(45^circ - theta))/(tan(60^circ + theta) * tan(30^circ - theta))`
= `(cos^2(45^circ + theta) + [sin{90^circ - (45^circ - theta)}]^2)/(tan(60^circ + theta) * cot{90^circ - (30^circ - theta)})` ...[∵ sin(90° – θ) = cos θ and cot(90° – θ) = tan θ]
= `(cos^2(45^circ + theta) + sin^2(45^circ + theta))/(tan(60^circ + theta) * cot(60^circ + theta))` ...[∵ sin2θ + cos2θ = 1]
= `1/(tan(60^circ + theta) * 1/(tan(60^circ + theta))` ...`[∵ cot θ = 1/tanθ]`
= 1
= R.H.S
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