Is tan(x)cot(x)=sec(x)csc(x) ?

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

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prove tan(x)cot(x)=sec(x)csc(x)

prove

tan (x) cot (x) = sec (x) csc (x)

False

Solution steps

tan (x) cot (x) = sec (x) csc (x)

Show that the two sides are not equal

For

tan (x) cot (x) = sec (x) csc (x)

plug in

x = 1

tan (1) cot (1) = 1

tan (1) cot (1)

Rewrite using trig identities:

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

cot (1)

Use the basic trigonometric identity:

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

= \frac{1}{tan ( 1 )}

= tan (1) \frac{1}{tan ( 1 )}

tan (1) \frac{1}{tan ( 1 )} = 1

tan (1) \frac{1}{tan ( 1 )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot tan ( 1 )}{tan ( 1 )}

Cancel the common factor:

tan (1)

= 1

= 1

sec (1) csc (1) = 2.19950 \dots

sec (1) csc (1)

Refine to a decimal form

= 2.19950 \dots

The two sides are not equal

\Rightarrow False

\frac{cos ( 3 x ) - cos ( 7 x )}{sin ( 3 x ) + sin ( 7 x )} = tan (2 x)

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

1 + tan (x) = sec (x)

2 sin^{2} (θ) - 1 = sin^{2} (θ) - cos^{2} (θ)

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

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