Is tan^2(x)-sin^2(x)=(tan(x)sin(x))^2 ?

The answer to whether tan^2(x)-sin^2(x)=(tan(x)sin(x))^2 is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove tan^2(x)-sin^2(x)=(tan(x)sin(x))^2

prove

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

True

Solution steps

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

Manipulating right side

(tan (x) sin (x))^{2}

Simplify

= sin^{2} (x) tan^{2} (x)

Rewrite using trig identities

sin^{2} (x) tan^{2} (x)

Use the Pythagorean identity:

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

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

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

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

Expand

(1 - cos^{2} (x)) tan^{2} (x) : tan^{2} (x) - tan^{2} (x) cos^{2} (x)

(1 - cos^{2} (x)) tan^{2} (x)

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

Apply the distributive law:

a (b - c) = ab - a c

a = tan^{2} (x), b = 1, c = cos^{2} (x)

= tan^{2} (x) \cdot 1 - tan^{2} (x) cos^{2} (x)

= 1 \cdot tan^{2} (x) - tan^{2} (x) cos^{2} (x)

Multiply:

1 \cdot tan^{2} (x) = tan^{2} (x)

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

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

Rewrite using trig identities

tan^{2} (x) - cos^{2} (x) tan^{2} (x)

Use the basic trigonometric identity:

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

= tan^{2} (x) - cos^{2} (x) (\frac{sin ( x )}{cos ( x )})^{2}

Simplify

cos^{2} (x) (\frac{sin ( x )}{cos ( x )})^{2} : sin^{2} (x)

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

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

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

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (x)

= sin^{2} (x)

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

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

We showed that the two sides could take the same form

\Rightarrow True

cos^{6} (10 5^{\circ}) = (\frac{1 + cos ( 21 0 ^{\circ} )}{2})^{3}

cos^{2} (θ) = sin^{2} (θ) - 1

\frac{tan ( 2 θ ) + cot ( 2 θ )}{sec ( 2 θ )} = csc (2 θ)

\frac{sec ( θ )}{sec ( θ ) - 1} = \frac{1}{1 - cos ( θ )}

2 cos (θ) - \frac{cos ( 2 θ )}{cos ( θ )} = sec (θ)

Is tan^2(x)-sin^2(x)=(tan(x)sin(x))^2 ?
The answer to whether tan^2(x)-sin^2(x)=(tan(x)sin(x))^2 is True