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

The answer to whether tan^2(y)-sin^2(y)=tan^2(y)sin^2(y) 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(y)-sin^2(y)=tan^2(y)sin^2(y)

prove

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

True

Solution steps

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

Manipulating right side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Expand

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

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

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

Multiply:

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (y)

= sin^{2} (y)

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

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

We showed that the two sides could take the same form

\Rightarrow True

sin (θ) (1 + cot^{2} (θ)) = csc (θ)

csc (x) - sin (x) = csc (x) cot (x)

5 + cot^{2} (x) = 4 + csc^{2} (x)

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

sin^{2} (2 x) = \frac{1 - cos ( 4 x )}{2}

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