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

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

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

prove

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

True

Solution steps

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

Manipulating left side

1 + tan^{2} (x)

Express with sin, cos

1 + tan^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

= 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 )}

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

cos (B) csc (B) tan (B) = 1

\frac{tan ( θ ) + cot ( θ )}{tan ( θ )} = csc^{2} (θ)

sin (6 0^{\circ}) = 2 sin (3 0^{\circ}) cos (3 0^{\circ})

\frac{cos ( θ )}{1 + sin ( θ )} + tan (θ) = sec (θ)

cos (x) \cdot tan (x) = sin (x)

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