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

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

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

prove

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

True

Solution steps

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

Manipulating right side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

cos (x) (tan (x) - sec (- x)) = sin (x) - 1

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

cot (u) - csc (u) = - \frac{sin ( u )}{1 + cos ( u )}

sec^{2} (θ) - tan^{2} (θ) = 1

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

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