$2 c o s^{2} A - 1 = 1 - 2 s i n^{2} A$
If true then write 1 and if false then write 0

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Solution

$A = 30^{\circ}$
Now, $2 {cos}^{2} A - 1 = 1 - 2 {sin}^{2} A$
L.H.S: $2 c o s^{2} A - 1 = 2 c o s^{2} 30^{0} - 1$
= $2 (\frac{\sqrt{3}}{2})^{2} - 1$
= $2 (\frac{3}{4}) - 1$
= $\frac{1}{2}$
R.H.S : $1 - 2 {sin}^{2} A$
= $1 - 2 {sin}^{2} 30^{0}$
= $1 - 2 (\frac{1}{2})^{2}$
= $1 - \frac{1}{2}$
= $\frac{1}{2}$
Thus, $L . H . S = R . H . S$

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A = 30^{0}

2 {cos}^{2} A - 1 = 1 - 2 {sin}^{2} A

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t a n (2 \times 30^{\circ}) = \frac{2 t a n 30^{\circ}}{1 - t a n^{2} 30^{\circ}}

s i n (2 \times 30^{\circ}) = \frac{2 t a n 30^{\circ}}{1 + t a n^{2} 30^{\circ}}

\frac{1 + s i n 2 A + c o s 2 A}{s i n A + c o s A} = 2 c o s A

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