Is sin^2(θ)+cos^2(θ)=sec^2(θ) ?

The answer to whether sin^2(θ)+cos^2(θ)=sec^2(θ) is False

prove sin^2(θ)+cos^2(θ)=sec^2(θ)

prove

sin^{2} (θ) + cos^{2} (θ) = sec^{2} (θ)

False

Solution steps

sin^{2} (θ) + cos^{2} (θ) = sec^{2} (θ)

Show that the two sides are not equal

For

sin^{2} (θ) + cos^{2} (θ) = sec^{2} (θ)

plug in

θ = 1

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

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

Use the Pythagorean identity:

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

= 1

= 1

sec^{2} (1) = 3.42551 \dots

sec^{2} (1)

Refine to a decimal form

= 3.42551 \dots

The two sides are not equal

\Rightarrow False

p ro v e sin (x) - cos (x) + 1 = 0

p ro v e csc^{4} (x) = csc^{3} (x)

p ro v e tan^{2} (x) = sec^{2} (x) - sec (x) tan (x) + 1

p ro v e \frac{sin ( 6 0 ^{\circ} )}{1 - cos ( 6 0 ^{\circ} )} = cot (3 0^{\circ})

p ro v e sin (2 x) + sin (x) = 0

Is sin^2(θ)+cos^2(θ)=sec^2(θ) ?
The answer to whether sin^2(θ)+cos^2(θ)=sec^2(θ) is False