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

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

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

\frac{1}{cos ( 0 ) + 1} + \frac{1}{tan ( 0 ) + 1} = \frac{3}{2} (Decimal : 1.5)

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

Use the following trivial identity:

tan (0) = 0

tan (0)

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 0

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

Simplify

\frac{1}{1 + 1} + \frac{1}{0 + 1} : \frac{3}{2}

\frac{1}{1 + 1} + \frac{1}{0 + 1}

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

\frac{1}{1 + 1}

Add the numbers:

1 + 1 = 2

= \frac{1}{2}

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

\frac{1}{0 + 1}

Add the numbers:

0 + 1 = 1

= \frac{1}{1}

Apply rule

\frac{a}{1} = a

= 1

= \frac{1}{2} + 1

Convert element to fraction:

1 = \frac{1 \cdot 2}{2}

= \frac{1 \cdot 2}{2} + \frac{1}{2}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 \cdot 2 + 1}{2}

1 \cdot 2 + 1 = 3

1 \cdot 2 + 1

Multiply the numbers:

1 \cdot 2 = 2

= 2 + 1

Add the numbers:

2 + 1 = 3

= 3

= \frac{3}{2}

= \frac{3}{2}

The two sides are not equal

\Rightarrow False

\frac{1}{csc ( x ) + 1} + \frac{1}{csc ( x ) - 1} = 2 sec (x) tan (x)

25 cos^{2} (x) = 25 - 25 sin^{2} (x)

\frac{1}{1 - sin ( θ )} + \frac{1}{1 + sin ( θ )} = 4 tan (θ)

sin (18 0^{\circ} - θ) = sin (θ)

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

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