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

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

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

prove

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

True

Solution steps

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

Manipulating left side

cos (x) (tan^{2} (x) + 1)

Express with sin, cos

(1 + tan^{2} (x)) cos (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Join

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

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

Multiply:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{s e c ( x )}}

Simplify

\frac{1}{\frac{1}{s e c ( x )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

Apply rule

\frac{a}{1} = a

= sec (x)

sec (x)

sec (x)

We showed that the two sides could take the same form

\Rightarrow True

csc^{2} (θ) - cot^{2} (θ) = 1

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

sin (θ) csc (θ) = 1

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

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

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