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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply exponent rule:

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (x)

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

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

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

Manipulating right side

2 - sin^{2} (x)

Rewrite using trig identities

2 - sin^{2} (x)

Use the Pythagorean identity:

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

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

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

Simplify

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

Subtract the numbers:

2 - 1 = 1

= cos^{2} (x) + 1

= cos^{2} (x) + 1

Use the Pythagorean identity:

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

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

Add similar elements:

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

\frac{sec ( θ ) cos ( θ )}{cot ( θ )} = tan (θ)

sin^{2} (θ - α) = 1

sin (x) (1 + cos (2 x)) = sin (2 x) cos (x)

(sec (x) - tan (x)) (1 + sin (x)) = cos (x)

tan (- x) = cos (x)

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