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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

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

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

\frac{tan ( β )}{sec ( β )} = sin (β)

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

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

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

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