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

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

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Popular >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= sin^{2} (x) + sin^{2} (x) (sec^{2} (x) - 1)

Simplify

sin^{2} (x) + sin^{2} (x) (sec^{2} (x) - 1) : sin^{2} (x) sec^{2} (x)

sin^{2} (x) + sin^{2} (x) (sec^{2} (x) - 1)

Expand

sin^{2} (x) (sec^{2} (x) - 1) : sin^{2} (x) sec^{2} (x) - sin^{2} (x)

sin^{2} (x) (sec^{2} (x) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = sin^{2} (x), b = sec^{2} (x), c = 1

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

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

Multiply:

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

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

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

Add similar elements:

sin^{2} (x) - sin^{2} (x) = 0

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

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

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

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Frequently Asked Questions (FAQ)

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