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Q.

x=a2cos2⁡α+b2sin2⁡α+a2sin2⁡α+b2cos2⁡α then x2=a2+b2+2pa2+b2−p2 , where p is equal to

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a

a2cos2⁡α+b2sin2⁡α

b

a2sin2⁡α+b2cos2⁡α

c

12a2+b2+a2−b2cos⁡2α

d

12a2+b2−a2−b2cos⁡2α

answer is A.

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Detailed Solution

x=a2cos2⁡α+b2sin2⁡α+a2sin2⁡α+b2cos2⁡α⇒x2=a2+b2+2a2cos2⁡α+b2sin2⁡αa2sin2⁡α+b2cos2⁡α= a2+b2+2k,where k= a2+b2−a2sin2⁡α+b2cos2⁡α×a2sin2⁡α+b2cos2⁡α∴x=a2+b2+2a2+b2p−p2where p=a2sin2⁡α+b2cos2⁡α=a22(1−cos⁡2α)+b22(1+cos⁡2α)=12a2+b2−a2−b2cos⁡2αAlso, x2=a2+b2+2a2cos2⁡α+b2sin2⁡αa2sin2⁡α+b2cos2⁡α=a2+b2+2a2+b2p−p2where, p=a2cos2⁡α+b2sin2⁡αor p=12a2+b2+a2−b2cos⁡2α
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