Is 1/(1-sin(θ))+1/(1+sin(θ))=4tan(θ) ?

The answer to whether 1/(1-sin(θ))+1/(1+sin(θ))=4tan(θ) is False

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prove 1/(1-sin(θ))+1/(1+sin(θ))=4tan(θ)

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

θ = 0

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

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

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

Simplify

= 2

4 tan (0) = 0

4 tan (0)

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

= 4 \cdot 0

Simplify

= 0

The two sides are not equal

\Rightarrow False

sin (18 0^{\circ} - θ) = sin (θ)

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

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

(cos (x) + sin (x)) 2 = 1 + 2 sin (x) cos (x)

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

Is 1/(1-sin(θ))+1/(1+sin(θ))=4tan(θ) ?
The answer to whether 1/(1-sin(θ))+1/(1+sin(θ))=4tan(θ) is False