Is 2-sec^2(θ)=1-tan^2(θ) ?

The answer to whether 2-sec^2(θ)=1-tan^2(θ) is True

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prove 2-sec^2(θ)=1-tan^2(θ)

prove

2 - sec^{2} (θ) = 1 - tan^{2} (θ)

True

Solution steps

2 - sec^{2} (θ) = 1 - tan^{2} (θ)

Manipulating left side

2 - sec^{2} (θ)

Rewrite using trig identities

2 - sec^{2} (θ)

Use the Pythagorean identity:

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

= 2 - (tan^{2} (θ) + 1)

Simplify

2 - (tan^{2} (θ) + 1) : - tan^{2} (θ) + 1

2 - (tan^{2} (θ) + 1)

- (tan^{2} (θ) + 1) : - tan^{2} (θ) - 1

- (tan^{2} (θ) + 1)

Distribute parentheses

= - (tan^{2} (θ)) - (1)

Apply minus-plus rules

+ (- a) = - a

= - tan^{2} (θ) - 1

= 2 - tan^{2} (θ) - 1

Simplify

2 - tan^{2} (θ) - 1 : - tan^{2} (θ) + 1

2 - tan^{2} (θ) - 1

Group like terms

= - tan^{2} (θ) + 2 - 1

Add/Subtract the numbers:

2 - 1 = 1

= - tan^{2} (θ) + 1

= - tan^{2} (θ) + 1

= - tan^{2} (θ) + 1

= - tan^{2} (θ) + 1

= 1 - tan^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{tan ( θ ) + sec ^{2} ( θ ) - sec ( θ )}{sec ( θ )} = tan^{2} (θ) + sin (θ)

sin (\frac{9 π}{4}) = sin (\frac{3}{4} π + 2 π)

\frac{cos ( 2 θ )}{sin ( 2 θ )} = 2 cos (2 θ)

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

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