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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Expand

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

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

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

Apply the distributive law:

a (b - c) = ab - a c

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

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

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

Multiply:

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (t)

= sin^{2} (t)

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

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

We showed that the two sides could take the same form

\Rightarrow True

8 sin^{2} (θ) + 9 cos^{2} (θ) = 8 + cos^{2} (θ)

2 = cos^{4} (+ sin^{4} (x))

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

\frac{sin ( a + b )}{sin ( a - b )} = \frac{1 + cot ( a ) tan ( b )}{1 - cot ( a ) tan ( b )}

cot (x + π) = cot (x)

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