Question

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: Question in attachment⤴

Answer 1

     

$(\sin\theta+\cos\theta)^2+(\sin\theta-\cos\theta)^2 \\ \\ 2(\sin^2\theta+\cos^2\theta)\ \ \ \ \ \ \ \ \ \ [(a+b)^2+(a-b)^2=2(a^2+b^2)] \\ \\ 2 \times 1 = \bold{2}$

$\therefore\ (\sin\theta+\cos\theta)^2+(\sin\theta-\cos\theta)^2=\bold{2}$

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Answer 2

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