Is 1+tan^2(A)=sec^3(A)cos(A) ?

The answer to whether 1+tan^2(A)=sec^3(A)cos(A) is True

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prove 1+tan^2(A)=sec^3(A)cos(A)

prove

1 + tan^{2} (A) = sec^{3} (A) cos (A)

True

Solution steps

1 + tan^{2} (A) = sec^{3} (A) cos (A)

Manipulating left side

1 + tan^{2} (A)

Express with sin, cos

1 + tan^{2} (A)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Convert element to fraction:

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

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

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} ( A ) + sin ^{2} ( A )}{cos ^{2} ( A )}

Multiply:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Manipulating right side

sec^{3} (A) cos (A)

Express with sin, cos

cos (A) sec^{3} (A)

Use the basic trigonometric identity:

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

= cos (A) (\frac{1}{cos ( A )})^{3}

Simplify

cos (A) (\frac{1}{cos ( A )})^{3} : \frac{1}{cos ^{2} ( A )}

cos (A) (\frac{1}{cos ( A )})^{3}

(\frac{1}{cos ( A )})^{3} = \frac{1}{cos ^{3} ( A )}

(\frac{1}{cos ( A )})^{3}

Apply exponent rule:

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

= \frac{1 ^{3}}{cos ^{3} ( A )}

Apply rule

1^{a} = 1

1^{3} = 1

= \frac{1}{cos ^{3} ( A )}

= \frac{1}{cos ^{3} ( A )} cos (A)

Multiply fractions:

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

= \frac{1 \cdot cos ( A )}{cos ^{3} ( A )}

Multiply:

1 \cdot cos (A) = cos (A)

= \frac{cos ( A )}{cos ^{3} ( A )}

Cancel the common factor:

cos (A)

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

sec (0) - cos (0) = tan (0) sin (0)

sin (x) - csc (x) = - cot (x) cos (x)

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

tan (x) \cdot cos (x) \cdot csc (x) = 1

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

Is 1+tan^2(A)=sec^3(A)cos(A) ?
The answer to whether 1+tan^2(A)=sec^3(A)cos(A) is True