Is sin^2(θ)+sin^2(θ)tan^2(θ)=tan^2(θ) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove sin^2(θ)+sin^2(θ)tan^2(θ)=tan^2(θ)

prove

sin^{2} (θ) + sin^{2} (θ) tan^{2} (θ) = tan^{2} (θ)

True

Solution steps

sin^{2} (θ) + sin^{2} (θ) tan^{2} (θ) = tan^{2} (θ)

Manipulating left side

sin^{2} (θ) + sin^{2} (θ) tan^{2} (θ)

Factor

sin^{2} (θ) + sin^{2} (θ) tan^{2} (θ) : sin^{2} (θ) (1 + tan^{2} (θ))

sin^{2} (θ) + sin^{2} (θ) tan^{2} (θ)

Factor out common term

sin^{2} (θ)

= sin^{2} (θ) (1 + tan^{2} (θ))

= (1 + tan^{2} (θ)) sin^{2} (θ)

Rewrite using trig identities

(1 + tan^{2} (θ)) sin^{2} (θ)

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

= sec^{2} (θ) sin^{2} (θ)

= sec^{2} (θ) sin^{2} (θ)

Express with sin, cos

sec^{2} (θ) sin^{2} (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

(\frac{1}{cos ( θ )})^{2} sin^{2} (θ)

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

(\frac{1}{cos ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{cos ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

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

Multiply fractions:

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

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

Multiply:

1 \cdot sin^{2} (θ) = sin^{2} (θ)

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

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

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

Rewrite using trig identities

= \frac{sin ( θ )}{cos ( θ )} \cdot \frac{sin ( θ )}{cos ( θ )}

Use the basic trigonometric identity:

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

\frac{sin ( θ ) tan ( θ )}{cos ( θ )}

= tan (θ) \frac{sin ( θ )}{cos ( θ )}

Use the basic trigonometric identity:

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

tan (θ) tan (θ)

Simplify

tan (θ) tan (θ) : tan^{2} (θ)

tan (θ) tan (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

tan (θ) tan (θ) = tan^{1 + 1} (θ)

= tan^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= tan^{2} (θ)

tan^{2} (θ)

tan^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

1 + sin (2 θ) = (cos (θ) sin (θ))^{2}

sec (θ) csc (θ) = tan (θ) + cot (θ)

2 tan (0^{\circ}) = sin (0^{\circ})

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

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

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