Is (cos(θ))/(1+sin(θ))=sec(θ)tan(θ) ?

The answer to whether (cos(θ))/(1+sin(θ))=sec(θ)tan(θ) is False

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prove (cos(θ))/(1+sin(θ))=sec(θ)tan(θ)

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

θ = 0

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

\frac{cos ( 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

= \frac{1}{1 + 0}

Simplify

= 1

sec (0) tan (0) = 0

sec (0) tan (0)

Use the following trivial identity:

sec (0) = 1

sec (0)

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

= 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

= 1 \cdot 0

Simplify

= 0

The two sides are not equal

\Rightarrow False

\frac{tan ( X ) - sin ( X )}{sin ^{3} ( X )} = \frac{sec ( X )}{1 + cos ( X )}

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

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

sin^{2} (t) = 2 sin (t) cos (t)

sin (\frac{2 π}{3}) = sin (\frac{- 11 π}{3})

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