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Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\rm \: If \: A = 2 {sin}^{2}\theta - cos2\theta \: then \: value \: of \: A \: lies \: in \: interval$

Solution :-

Identities Used :-

$\boxed{ \red{ \sf \:cos2x = 1 - {2sin}^{2} x}}$

$\boxed{ \red{ \sf \: - 1 \leqslant sinx \leqslant 1}}$

Let's solve the problem now!!

Consider,

$\rm :\longmapsto\:\: A = 2 {sin}^{2}\theta - cos2\theta$

$\rm :\longmapsto\:\: A = 2 {sin}^{2}\theta - (1 - 2 {sin}^{2} \theta)$

$\rm :\longmapsto\:\: A = 2 {sin}^{2}\theta - 1 + 2 {sin}^{2} \theta$

$\rm :\longmapsto\:\: A = 4 {sin}^{2}\theta - 1$

We know that,

$\rm :\longmapsto\: - 1 \leqslant sin\theta \leqslant 1$

On squaring, we get

$\rm :\implies\:0 \leqslant {sin}^{2} \theta \leqslant 1$

Multiply by 4 each term, we get

$\rm :\longmapsto\:0 \leqslant 4 {sin}^{2} \theta \leqslant 4$

On Subtracting 1 from each term, we get

$\rm :\longmapsto\: 0 - 1 \leqslant 4 {sin}^{2}\theta - 1 \leqslant 4 - 1$

$\rm :\longmapsto\: - 1 \leqslant 4 {sin}^{2}\theta - 1 \leqslant 3$

$\bf\implies \: - 1 \leqslant A \leqslant 3$

$\boxed{ \red{ \bf \:Hence, \: option \: (a) \: is \: correct.}}$

Additional Information :-

$\boxed{ \red{ \sf \: - 1 \leqslant cos\theta \leqslant 1}}$

$\boxed{ \red{ \sf \:cos2x = {cos}^{2} x - {sin}^{2}x}}$

$\boxed{ \red{ \sf \:cos2x = 2 {cos}^{2} x - 1}}$

$\boxed{ \red{ \sf \:sin2x = 2sinxcosx}}$