Is cos^2(α)+cos^2(α)tan^2(α)=1 ?

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

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prove cos^2(α)+cos^2(α)tan^2(α)=1

prove

cos^{2} (α) + cos^{2} (α) tan^{2} (α) = 1

True

Solution steps

cos^{2} (α) + cos^{2} (α) tan^{2} (α) = 1

Manipulating left side

cos^{2} (α) + cos^{2} (α) tan^{2} (α)

Express with sin, cos

cos^{2} (α) + cos^{2} (α) tan^{2} (α)

Use the basic trigonometric identity:

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

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

Simplify

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

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

cos^{2} (α) (\frac{sin ( α )}{cos ( α )})^{2} = sin^{2} (α)

cos^{2} (α) (\frac{sin ( α )}{cos ( α )})^{2}

(\frac{sin ( α )}{cos ( α )})^{2} = \frac{sin ^{2} ( α )}{cos ^{2} ( α )}

(\frac{sin ( α )}{cos ( α )})^{2}

Apply exponent rule:

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

= \frac{sin ^{2} ( α )}{cos ^{2} ( α )}

= \frac{sin ^{2} ( α )}{cos ^{2} ( α )} cos^{2} (α)

Multiply fractions:

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

= \frac{sin ^{2} ( α ) cos ^{2} ( α )}{cos ^{2} ( α )}

Cancel the common factor:

cos^{2} (α)

= sin^{2} (α)

= cos^{2} (α) + sin^{2} (α)

= cos^{2} (α) + sin^{2} (α)

= cos^{2} (α) + sin^{2} (α)

Rewrite using trig identities

cos^{2} (α) + sin^{2} (α)

Use the Pythagorean identity:

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

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

cos (\frac{π}{7}) = sin (\frac{π}{2} - \frac{π}{7})

sin (\frac{π}{4} + x) = 2 (cos (x) + sin (x))

sec (x) csc (x) = cot (x) + tan (x)

arctan (x) = \frac{arcsin ( x )}{arccos ( x )}

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

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