Is sec^2(x)-tan^2(x)=tan(x)cot(x) ?

The answer to whether sec^2(x)-tan^2(x)=tan(x)cot(x) is True

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prove sec^2(x)-tan^2(x)=tan(x)cot(x)

prove

sec^{2} (x) - tan^{2} (x) = tan (x) cot (x)

True

Solution steps

sec^{2} (x) - tan^{2} (x) = tan (x) cot (x)

Manipulating right side

tan (x) cot (x)

Express with sin, cos

cot (x) tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

\frac{cos ( x )}{sin ( x )} \cdot \frac{sin ( x )}{cos ( x )} = 1

\frac{cos ( x )}{sin ( x )} \cdot \frac{sin ( x )}{cos ( x )}

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

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

Cancel the common factor:

sin (x)

= 1

= 1

= 1

Rewrite using trig identities

1

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

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

= sec^{2} (x) - tan^{2} (x)

= sec^{2} (x) - tan^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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

csc (u) - sin (u) = cos (u) cot (u)

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

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

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

Is sec^2(x)-tan^2(x)=tan(x)cot(x) ?
The answer to whether sec^2(x)-tan^2(x)=tan(x)cot(x) is True