Is tan^2(x)=sec(x)csc(x)tan(x)-1 ?

The answer to whether tan^2(x)=sec(x)csc(x)tan(x)-1 is True

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prove tan^2(x)=sec(x)csc(x)tan(x)-1

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

sec (x) csc (x) tan (x) - 1

Express with sin, cos

- 1 + csc (x) sec (x) tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

- 1 + \frac{1}{sin ( x )} \cdot \frac{1}{cos ( x )} \cdot \frac{sin ( x )}{cos ( x )} : \frac{- cos ^{2} ( x ) + 1}{cos ^{2} ( x )}

- 1 + \frac{1}{sin ( x )} \cdot \frac{1}{cos ( x )} \cdot \frac{sin ( x )}{cos ( x )}

\frac{1}{sin ( x )} \cdot \frac{1}{cos ( x )} \cdot \frac{sin ( x )}{cos ( x )} = \frac{1}{cos ^{2} ( x )}

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

Multiply fractions:

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

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

Cancel the common factor:

sin (x)

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

Multiply the numbers:

1 \cdot 1 = 1

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

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

cos (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (x) cos (x) = cos^{1 + 1} (x)

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

= \frac{1}{cos ^{2} ( x )}

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

Convert element to fraction:

1 = \frac{1 cos ^{2} ( x )}{cos ^{2} ( x )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

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

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

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

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

Rewrite using trig identities

\frac{1 - cos ^{2} ( x )}{cos ^{2} ( x )}

Use the Pythagorean identity:

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

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

Use the basic trigonometric identity:

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

tan (x) tan (x)

Simplify

tan (x) tan (x) : tan^{2} (x)

tan (x) tan (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= tan^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= tan^{2} (x)

tan^{2} (x)

tan^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is tan^2(x)=sec(x)csc(x)tan(x)-1 ?
The answer to whether tan^2(x)=sec(x)csc(x)tan(x)-1 is True