Is tan(θ)cos(θ)-sin^2(θ)=cos^2(θ) ?

The answer to whether tan(θ)cos(θ)-sin^2(θ)=cos^2(θ) is False

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prove tan(θ)cos(θ)-sin^2(θ)=cos^2(θ)

prove

tan (θ) cos (θ) - sin^{2} (θ) = cos^{2} (θ)

False

Solution steps

tan (θ) cos (θ) - sin^{2} (θ) = cos^{2} (θ)

Show that the two sides are not equal

For

tan (θ) cos (θ) - sin^{2} (θ) = cos^{2} (θ)

plug in

θ = 0

tan (0) cos (0) - sin^{2} (0) = 0

tan (0) cos (0) - sin^{2} (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

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

= 0 \cdot 1 - 0^{2}

Simplify

= 0

cos^{2} (0) = 1

cos^{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

= 1^{2}

Simplify

= 1

The two sides are not equal

\Rightarrow False

\frac{cot ^{2} ( θ )}{cos ^{2} ( θ )} = csc^{2} (θ)

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

\frac{1 + sec ( θ )}{sec ^{2} ( θ )} = 1 + cos^{2} (θ)

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

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

Is tan(θ)cos(θ)-sin^2(θ)=cos^2(θ) ?
The answer to whether tan(θ)cos(θ)-sin^2(θ)=cos^2(θ) is False