Is (1-cos^2(a))(1+tan^2(a))=tan^2(a) ?

The answer to whether (1-cos^2(a))(1+tan^2(a))=tan^2(a) is True

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prove (1-cos^2(a))(1+tan^2(a))=tan^2(a)

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Express with sin, cos

(1 - cos^{2} (a)) sec^{2} (a)

Use the basic trigonometric identity:

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

= (1 - cos^{2} (a)) (\frac{1}{cos ( a )})^{2}

Simplify

(1 - cos^{2} (a)) (\frac{1}{cos ( a )})^{2} : \frac{1 - cos ^{2} ( a )}{cos ^{2} ( a )}

(1 - cos^{2} (a)) (\frac{1}{cos ( a )})^{2}

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= \frac{1}{cos ^{2} ( a )} (- cos^{2} (a) + 1)

Multiply fractions:

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

= \frac{1 \cdot ( 1 - cos ^{2} ( a ) )}{cos ^{2} ( a )}

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

1 \cdot (1 - cos^{2} (a))

Multiply:

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

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

Remove parentheses:

(a) = a

= 1 - cos^{2} (a)

= \frac{1 - cos ^{2} ( a )}{cos ^{2} ( a )}

= \frac{1 - cos ^{2} ( a )}{cos ^{2} ( a )}

= \frac{1 - cos ^{2} ( a )}{cos ^{2} ( a )}

Rewrite using trig identities

\frac{1 - cos ^{2} ( a )}{cos ^{2} ( a )}

Use the Pythagorean identity:

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

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

Use the basic trigonometric identity:

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

tan (a) tan (a)

Simplify

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

tan (a) tan (a)

Apply exponent rule:

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

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

= tan^{1 + 1} (a)

Add the numbers:

1 + 1 = 2

= tan^{2} (a)

tan^{2} (a)

tan^{2} (a)

We showed that the two sides could take the same form

\Rightarrow True

\frac{1}{sin ( \frac{13}{5} )} = csc (\frac{13}{5})

tan (\frac{θ}{2}) = \frac{1 - cos ( θ )}{2}

- csc (x) + sin (x) = - cos (x) cot (x)

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

cot (x) sec (x) csc^{2} (x) - cot^{3} (x) sec (x) = csc (x)

Is (1-cos^2(a))(1+tan^2(a))=tan^2(a) ?
The answer to whether (1-cos^2(a))(1+tan^2(a))=tan^2(a) is True