Is tan^2(15)-sec^2(15)=-1 ?

The answer to whether tan^2(15)-sec^2(15)=-1 is True

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prove tan^2(15)-sec^2(15)=-1

prove

tan^{2} (1 5^{\circ}) - sec^{2} (1 5^{\circ}) = - 1

True

Solution steps

tan^{2} (1 5^{\circ}) - sec^{2} (1 5^{\circ}) = - 1

Use the Pythagorean identity:

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

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

\Rightarrow True

cos (2 a) = \frac{cot ( a ) - tan ( a )}{csc ( a ) sec ( a )}

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

cos (a) cos (b) = \frac{1}{2} (cos (a - b) + cos (a + b))

sin^{2} (θ) (cot^{2} (θ) + 1) = 1

sin (a - b) = sin (a) cos (b) - cos (a) sin (b)

Is tan^2(15)-sec^2(15)=-1 ?
The answer to whether tan^2(15)-sec^2(15)=-1 is True