Is 1+(cos^2(x))/(1+sin(x))=sin^2(x) ?

The answer to whether 1+(cos^2(x))/(1+sin(x))=sin^2(x) is False

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prove 1+(cos^2(x))/(1+sin(x))=sin^2(x)

prove

1 + \frac{cos ^{2} ( x )}{1 + sin ( x )} = sin^{2} (x)

False

Solution steps

1 + \frac{cos ^{2} ( x )}{1 + sin ( x )} = sin^{2} (x)

Show that the two sides are not equal

For

1 + \frac{cos ^{2} ( x )}{1 + sin ( x )} = sin^{2} (x)

plug in

x = 0

1 + \frac{cos ^{2} ( 0 )}{1 + sin ( 0 )} = 2

1 + \frac{cos ^{2} ( 0 )}{1 + sin ( 0 )}

Use the following trivial identity:

cos (0) = 1

cos (0)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 1

Use the following trivial identity:

sin (0) = 0

sin (0)

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= 1 + \frac{1 ^{2}}{1 + 0}

Simplify

= 2

sin^{2} (0) = 0

sin^{2} (0)

Use the following trivial identity:

sin (0) = 0

sin (0)

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 0

= 0^{2}

Simplify

= 0

The two sides are not equal

\Rightarrow False

16 (\frac{1 - cos ( 2 θ )}{2}) = 8 - 8 cos (2 θ)

cot (θ) sin (θ) cos (θ) = cos^{2} (θ)

sec (x) + tan (x) = \frac{sin ( x ) + 1}{cos ( x )}

1 - \frac{2}{1 + cot ^{2} ( x )} = cos (2 x)

tan (x) - 1 = cot (x)

Is 1+(cos^2(x))/(1+sin(x))=sin^2(x) ?
The answer to whether 1+(cos^2(x))/(1+sin(x))=sin^2(x) is False