Is 1/(sec(θ)-tan(θ))=sec(θ)tan(θ) ?

The answer to whether 1/(sec(θ)-tan(θ))=sec(θ)tan(θ) is False

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prove 1/(sec(θ)-tan(θ))=sec(θ)tan(θ)

prove

\frac{1}{sec ( θ ) - tan ( θ )} = sec (θ) tan (θ)

False

Solution steps

\frac{1}{sec ( θ ) - tan ( θ )} = sec (θ) tan (θ)

Show that the two sides are not equal

For

\frac{1}{sec ( θ ) - tan ( θ )} = sec (θ) tan (θ)

plug in

θ = 0

\frac{1}{sec ( 0 ) - tan ( 0 )} = 1

\frac{1}{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

= \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

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

16 (\frac{1 - cos ( 2 θ )}{2}) = 8 - 8 cos (2 θ)

cot (θ) sin (θ) cos (θ) = cos^{2} (θ)

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

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

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