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

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

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

prove

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

True

Solution steps

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

Manipulating right side

sec^{2} (x) (1 - cos^{2} (x))

Express with sin, cos

(1 - cos^{2} (x)) sec^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Multiply fractions:

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

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

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

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

Multiply:

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

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

Remove parentheses:

(a) = a

= 1 - cos^{2} (x)

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

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

tan (x) = tan (x) csc^{2} (x) + cot (- x)

tan (8 x) = \frac{2 tan ( x )}{1 - tan ^{2} ( x )}

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

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

3 - 3 (cos (x) - sin (x))^{2} = - 3 cos (2 x)

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