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Question

If cos(θ-α)=a,sin(θ-β)=b, then a2-2absin(α-β)+b2 is equal to


A

sin2(αβ)

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B

cos2(αβ)

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C

1

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D

0

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Solution

The correct option is B

cos2(αβ)


Step 1. Find the value of a2-2absin(α-β)+b2:

Given, cos(θ-α)=a,sin(θ-β)=b

αβ=(θβ)(θα)

Step 2. Take “sin” on both sides,

sin(αβ)=sin[(θβ)(θα)]=sin(θβ)cos(θα)cos(θβ)sin(θα)=ba(1b2)(1a2)

sin(α-β)=ab(1a2)(1b2) ….(i)

Again,

αβ=(θβ)(θα)

Step 3. Take “cos” on both sides,

cos(αβ)=cos[(θβ)(θα)]=cos(θα)cos(θβ)+sin(θα)sin(θβ)=a(1b2)+b(1a2)

cos2(αβ)=a2(1b2)+b2(1a2)+2ab(1a2)(1b2) ….(ii)

a22absin(αβ)+b2=a22ab[ab(1a2)(1b2)]+b2 {from (i)}

=a22a2b2+2ab(1a2)(1b2)+b2 …..(iii)

Step 4. By equating equation (ii) and (iii), we get

a22absin(αβ)+b2=cos2(αβ)

Hence, Option ‘B’ is Correct.


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