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

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

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

sin (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

The two sides are not equal

\Rightarrow False

sec^{2} (x) (1 - csc^{2} (x)) = - csc^{2} (x)

- sin^{2} (x) = cos^{2} (x) - 1

cos (x) + sin (x) (tan (x)) = sec (x)

\frac{sin ( A )}{1 - cos ( A )} - cot (A) = csc (A)

(cos (2 x) - sin (2 x))^{2} - 1 = sin (- 4 x)

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