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

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

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

prove

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

True

Solution steps

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

Manipulating right side

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

Expand

(- 1 + cos^{2} (x)) (1 + tan^{2} (x)) : - 1 - tan^{2} (x) + cos^{2} (x) + cos^{2} (x) tan^{2} (x)

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

Apply FOIL method:

(a + b) (c + d) = a c + a d + b c + b d

a = - 1, b = cos^{2} (x), c = 1, d = tan^{2} (x)

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

Apply minus-plus rules

+ (- a) = - a

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

Simplify

- 1 \cdot 1 - 1 \cdot tan^{2} (x) + 1 \cdot cos^{2} (x) + cos^{2} (x) tan^{2} (x) : - 1 - tan^{2} (x) + cos^{2} (x) + cos^{2} (x) tan^{2} (x)

- 1 \cdot 1 - 1 \cdot tan^{2} (x) + 1 \cdot cos^{2} (x) + cos^{2} (x) tan^{2} (x)

Multiply the numbers:

1 \cdot 1 = 1

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

Multiply:

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

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

Multiply:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (x)

= sin^{2} (x)

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

Add similar elements:

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

= - tan^{2} (x)

= - tan^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

csc (x) tan (x) cos^{2} (x) = cos (x)

sin (2 x) cot (x) = 2 cos^{2} (x)

sin (a) sin (b) = sin (a + b)

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

2 sin (x) = \frac{( 4 cos ( x ) - 1 )}{tan ( x )}

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