1 Simplify the following equation without using a calculator: 2 sec 300° – tan²(-135°) – 2 cos² 90° + 3 sin 210°

• $\sec 300^\circ = \frac{1}{\cos 300^\circ} = \frac{1}{\cos(360^\circ - 60^\circ)} = \frac{1}{\cos 60^\circ} = \frac{1}{1/2} = 2$.
• $\tan(-135^\circ) = \tan(-135^\circ + 360^\circ) = \tan(225^\circ) = \tan(180^\circ + 45^\circ) = \tan 45^\circ = 1$. So, $\tan^2(-135^\circ) = (1)^2 = 1$.
• $\cos 90^\circ = 0$. So, $2 \cos^2 90^\circ = 2(0)^2 = 0$.
• $\sin 210^\circ = \sin(180^\circ + 30^\circ) = -\sin 30^\circ = -\frac{1}{2}$.

The simplified value is $\boxed{\frac{3}{2}}$.

$\sin(A+B) = \sin A \cos B + \cos A \sin B$.
Let $A=x$ and $B=45^\circ$.

We know that $\cos 45^\circ = \frac{\sqrt{2}}{2}$ and $\sin 45^\circ = \frac{\sqrt{2}}{2}$.

Since $\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$, we can write:

This matches the right-hand side, thus the identity is proven.

• For $n=0$: $x = 67.5^\circ - 0 \cdot 90^\circ = 67.5^\circ$. This value is within the range $[0^\circ, 90^\circ]$.
• For $n=1$: $x = 67.5^\circ - 1 \cdot 90^\circ = -22.5^\circ$. This value is not in the range.
• For $n=-1$: $x = 67.5^\circ - (-1) \cdot 90^\circ = 67.5^\circ + 90^\circ = 157.5^\circ$. This value is not in the range. The only value of $x$ in the specified range is $\boxed{67.5^\circ}$.

• $\cos\left(\frac{\pi}{2}-x\right) = \sin x$ (co-function identity).
• $\sin(\pi-y) = \sin y$ (sine is positive in the second quadrant). Substitute these into the denominator:

The formula for $\cos(x+y)$ is $\cos x \cos y - \sin x \sin y$.
Substitute this into the denominator:

The formula for $\sin(x+y)$ is $\sin x \cos y + \cos x \sin y$.
Substitute this into the RHS:

This matches the left-hand side (LHS) of the equation, thus the identity is proven.