Is (1+tan^2(x))(sin^2(x))=tan^2(x) ?

The answer to whether (1+tan^2(x))(sin^2(x))=tan^2(x) is True

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prove (1+tan^2(x))(sin^2(x))=tan^2(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Express with sin, cos

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

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

Simplify

(\frac{1}{cos ( x )})^{2} sin^{2} (x) : \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

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

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

(\frac{1}{cos ( x )})^{2}

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Multiply fractions:

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

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

Multiply:

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

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

Use the basic trigonometric identity:

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

tan (x) tan (x)

Simplify

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

tan (x) tan (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= tan^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= tan^{2} (x)

tan^{2} (x)

tan^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

tan (\frac{11}{12} π) = tan (- \frac{π}{12})

cot (B) sec (B) sin (B) = 1

\frac{cos ( x ) - tan ( x )}{sin ( x ) - cot ( x )} = - 1

1 - 2 sin^{2} (2 A) = 8 cos^{4} (A) - 8 cos^{2} (A) + 1

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

Is (1+tan^2(x))(sin^2(x))=tan^2(x) ?
The answer to whether (1+tan^2(x))(sin^2(x))=tan^2(x) is True