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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Expand

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

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

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

Multiply:

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (a)

= sin^{2} (a)

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

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

We showed that the two sides could take the same form

\Rightarrow True

\frac{cos ( a )}{1 - sin ( a )} - tan (a) = sec (a)

tan (θ) (sin (θ) + cot (θ) cos (θ)) = sec (θ)

sin (x + y) = sin (x) + sin (y)

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

sin^{2} (2 θ) = 4 sin^{2} (θ) cos^{2} (θ)

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