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

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

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Popular >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

sin^{2} (α) + sin^{2} (α) tan^{2} (α)

Rewrite using trig identities

sin^{2} (α) + sin^{2} (α) tan^{2} (α)

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

sin^{2} (α) (sec^{2} (α) - 1)

Apply the distributive law:

a (b - c) = ab - a c

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

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

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

Multiply:

1 \cdot sin^{2} (α) = sin^{2} (α)

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

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

Add similar elements:

sin^{2} (α) - sin^{2} (α) = 0

= sin^{2} (α) sec^{2} (α)

= sin^{2} (α) sec^{2} (α)

= sin^{2} (α) sec^{2} (α)

Express with sin, cos

sec^{2} (α) sin^{2} (α)

Use the basic trigonometric identity:

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

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

Simplify

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

(\frac{1}{cos ( α )})^{2} sin^{2} (α)

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

(\frac{1}{cos ( α )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{cos ^{2} ( α )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( α )}

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

Multiply fractions:

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

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

Multiply:

1 \cdot sin^{2} (α) = sin^{2} (α)

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

\frac{sin ( α ) tan ( α )}{cos ( α )}

= tan (α) \frac{sin ( α )}{cos ( α )}

Use the basic trigonometric identity:

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

tan (α) tan (α)

Simplify

tan (α) tan (α) : tan^{2} (α)

tan (α) tan (α)

Apply exponent rule:

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

tan (α) tan (α) = tan^{1 + 1} (α)

= tan^{1 + 1} (α)

Add the numbers:

1 + 1 = 2

= tan^{2} (α)

tan^{2} (α)

tan^{2} (α)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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