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

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

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

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 1

cot (1) tan (1) = 1

cot (1) tan (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 )}

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

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

\frac{1}{tan ( 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

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

csc (1) sec (1)

Refine to a decimal form

= 2.19950 \dots

The two sides are not equal

\Rightarrow False

sec (x) - cos (x) = (1 - cos (x)) (1 + sec (x))

(sec^{2} (x) - 1) (sin^{2} (x) - 1) = - sin^{2} (x)

csc^{2} (x) - csc^{2} (x) \cdot cos^{2} (x) = 1

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

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

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