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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

6 cos^{2} (x) - 6 sin^{2} (x) = 12 cos^{2} (x) - 6

sin (θ) cot (θ) sec (θ) = 1

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

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

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