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

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

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

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

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

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

= \frac{1 + 0}{1}

Simplify

= 1

cos (0) + 1 = 2

cos (0) + 1

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

= 1 + 1

Simplify

= 2

The two sides are not equal

\Rightarrow False

\frac{csc ( a )}{cot ( a )} = sec (a)

(tan (x) - 1) cos (x) = sin (x) - cos (x)

sin (θ - \frac{π}{2}) = - cos (θ)

sec (x) - cos (x) = tan (x) \cdot sin (x)

\frac{cot ( x ) - 1}{1 - tan ( x )} = cot (x)

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