Is cos^2(θ)tan^2(θ)+cos^2(θ)=1 ?

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

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prove cos^2(θ)tan^2(θ)+cos^2(θ)=1

prove

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

True

Solution steps

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

Manipulating left side

cos^{2} (θ) tan^{2} (θ) + cos^{2} (θ)

Express with sin, cos

cos^{2} (θ) + cos^{2} (θ) tan^{2} (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

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

(\frac{sin ( θ )}{cos ( θ )})^{2}

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (θ)

= sin^{2} (θ)

= cos^{2} (θ) + sin^{2} (θ)

= cos^{2} (θ) + sin^{2} (θ)

= cos^{2} (θ) + sin^{2} (θ)

Rewrite using trig identities

cos^{2} (θ) + sin^{2} (θ)

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

\frac{sin ( θ ) csc ( θ )}{tan ( θ )} = cot (θ)

sec^{2} (θ) + csc (θ) = - sec^{2} (θ) csc (θ)

tan (5 x) = tan (3 x) sec (2 x) - tan (3 x)

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

sin (4 0^{\circ}) = 2 \cdot sin (2 0^{\circ}) \cdot cos (2 0^{\circ})

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