Is (sin^2(θ))/(cos(θ))+cos(θ)=sec(θ) ?

The answer to whether (sin^2(θ))/(cos(θ))+cos(θ)=sec(θ) is True

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prove (sin^2(θ))/(cos(θ))+cos(θ)=sec(θ)

prove

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

True

Solution steps

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

Manipulating right side

sec (θ)

Express with sin, cos

sec (θ)

Use the basic trigonometric identity:

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

= \frac{1}{cos ( θ )}

= \frac{1}{cos ( θ )}

Rewrite using trig identities

\frac{1}{cos ( θ )}

Use the Pythagorean identity:

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

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

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

Expand

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

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

Apply the fraction rule:

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

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

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

Cancel

\frac{cos ^{2} ( θ )}{cos ( θ )} : cos (θ)

\frac{cos ^{2} ( θ )}{cos ( θ )}

Cancel the common factor:

cos (θ)

= cos (θ)

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

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

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

- \frac{1}{2} = sin (- \frac{π}{12}) 2 + 3

- sin (\frac{π}{4}) = - cos (\frac{π}{4})

Is (sin^2(θ))/(cos(θ))+cos(θ)=sec(θ) ?
The answer to whether (sin^2(θ))/(cos(θ))+cos(θ)=sec(θ) is True