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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

(1 - (\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 )} + 1)

Join

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

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

Convert element to fraction:

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

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

Multiply:

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

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

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

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

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

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

Cancel the common factor:

cos^{2} (x)

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

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

Add similar elements:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

csc^{3} (x) + csc^{2} (x) + csc (x) = csc (x)

1 - 8 sin^{2} (x) cos^{2} (x) = cos (4 x)

5 tan (2 t) + 3 = 3

(1 + sin (y)) (1 + sin (- y)) = cos^{2} (y)

tan (x) = - cot (2 x) + csc (2 x)

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