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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

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

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

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

= 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

cot (a) - 1 = \frac{csc ( a ) - sec ( a )}{sec ( a )}

tan (13 5^{\circ} + θ) = \frac{- 1 + tan ( θ )}{1 + tan ( θ )}

\frac{csc ( \frac{π}{2} - x )}{- tan ( - x )} = csc (x)

cot^{2} (θ) cos^{2} (θ) = cot^{2} (θ) - cos^{2} (θ)

tan (12 0^{\circ}) = tan (18 0^{\circ} - 6 0^{\circ})

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