Determine the value of the expression without the use of a culculator: \[ \frac{\tan 300^{\circ}+\cos \left(90^{\circ}+x\right)}{\sin \left(180^{\circ}-x\right)+2 \cos \left(-30^{\circ}\right)} \] Prove the following identity: \( \frac{1-\sin 2 x}{\cos 2 x}=\frac{\cos x-\sin x}{\cos x+\sin x} \) Determine the general solution of \( \cos x-\sin x=\sqrt{2} \)

We will solve three parts: ────────────────────────────── Part 1. Evaluate   A = [tan 300° + cos(90° + x)] ⁄ [sin(180° − x) + 2 cos(−30°)] Step 1. Compute tan 300°.  • Note that 300° = 360° − 60°, so   tan 300° = tan(−60°) = −tan 60° = −√3. Step 2. Compute cos(90° + x).  • Using the identity cos(90° + x) = −sin x, we get   cos(90° + x) = −sin x. Step 3. Compute sin(180° − x).  • Recall sin(180° − x) = sin x. Step 4. Compute cos(−30°).  • Since cosine is even, cos(−30°) = cos 30° = √3⁄2. Step 5. Write the numerator and denominator.  • Numerator: tan 300° + cos(90° + x) = (−√3) + (− sin x) = − (√3 + sin x).  • Denominator: sin(180° − x) + 2 cos(−30°)         = sin x + 2(√3⁄2) = sin x + √3. Step 6. Divide the two:   A = [−(√3 + sin x)] / (sin x + √3) = −1. Thus, the value of the expression is –1. ────────────────────────────── Part 2. Prove the identity   (1 − sin 2x)⁄(cos 2x) = (cos x − sin x)⁄(cos x + sin x) Proof: Step 1. Express sin 2x and cos 2x in terms of sin x and cos x.  • sin 2x = 2 sin x cos x,  • cos 2x = cos²x − sin²x. Step 2. Substitute into the left-hand side (LHS):   LHS = (1 − 2 sin x cos x)⁄(cos²x − sin²x). Step 3. Notice that the numerator can be written as:   1 − 2 sin x cos x = (sin²x + cos²x) − 2 sin x cos x = cos²x − 2 sin x cos x + sin²x = (cos x − sin x)². Step 4. Also, write the denominator as:   cos²x − sin²x = (cos x + sin x)(cos x − sin x). Step 5. Substitute these into LHS:   LHS = [(cos x − sin x)²]⁄[(cos x + sin x)(cos x − sin x)]. Provided cos x − sin x ≠ 0 (which is acceptable for the identity’s validity), one factor cancels:   LHS = (cos x − sin x)⁄(cos x + sin x). This is exactly the right-hand side (RHS). Hence, the identity is proven. ────────────────────────────── Part 3. Determine the general solution of   cos x − sin x = √2 Solution: Step 1. Recognize that the left side can be written in the form R cos(x + φ). In fact, note that   cos x − sin x = √2 cos(x + π⁄4) Why?  • Recall cos(x + π⁄4) = cos x cos(π⁄4) − sin x sin(π⁄4)  • And since cos(π⁄4) = sin(π⁄4) = √2⁄2, we have   cos(x + π⁄4) = (√2⁄2)(cos x − sin x).  • Multiply both sides by √2:   √2 cos(x + π⁄4) = cos x − sin x. Step 2. Substitute into the equation:   √2 cos(x + π⁄4) = √2. Step 3. Divide both sides by √2 (assuming √2 ≠ 0):   cos(x + π⁄4) = 1. Step 4. Solve cosθ = 1.   We have cosθ = 1 when θ = 2πk for any integer k. Here θ = x + π⁄4, so:   x + π⁄4 = 2πk  ⇒  x = 2πk − π⁄4. Thus, the general solution is:   x = 2πk − π⁄4,  where k is any integer. ────────────────────────────── Summary of Answers: 1. The expression evaluates to –1. 2. The identity (1 − sin 2x)/(cos 2x) = (cos x − sin x)/(cos x + sin x) is proven as shown above. 3. The general solution of cos x − sin x = √2 is x = 2πk − π⁄4 for any integer k.