Is cot(θ)+tan(θ)=sec(θ)+csc(θ) ?

The answer to whether cot(θ)+tan(θ)=sec(θ)+csc(θ) is False

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prove cot(θ)+tan(θ)=sec(θ)+csc(θ)

prove

cot (θ) + tan (θ) = sec (θ) + csc (θ)

False

Solution steps

cot (θ) + tan (θ) = sec (θ) + csc (θ)

Show that the two sides are not equal

For

cot (θ) + tan (θ) = sec (θ) + csc (θ)

plug in

θ = 1

cot (1) + tan (1) = 2.19950 \dots

cot (1) + tan (1)

Refine to a decimal form

= 2.19950 \dots

sec (1) + csc (1) = 3.03921 \dots

sec (1) + csc (1)

Refine to a decimal form

= 3.03921 \dots

The two sides are not equal

\Rightarrow False

sin (75 0^{\circ}) = sin (45 0^{\circ} + 30 0^{\circ})

2 sin (θ) + sin (2 θ) = 0

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

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

tan (x - \frac{3 π}{2}) = - cot (x)

Is cot(θ)+tan(θ)=sec(θ)+csc(θ) ?
The answer to whether cot(θ)+tan(θ)=sec(θ)+csc(θ) is False