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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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sin4A – cos4A = 1 – 2cos2A. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` .....`[sin^2"A" + square = 1]`
= `square` – cos2A .....[sin2A = 1 – cos2A]
= `square`
= R.H.S
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Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
If cos A + cos2A = 1, then sin2A + sin4 A = ?
sin(45° + θ) – cos(45° – θ) is equal to ______.
