$$ \cos \left(-60^{\circ}\right)-\tan 135 $$ 4.4 Prove that: $$ \frac{\cos x}{1-\sin x}-\frac{\cos x}{1+\sin x}=2 \tan x $$ 4.5 Determine the general solution of $$ \cos ^{2} x-3 \sin x-3=-3 \sin ^{2} x $$

Question

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1. Evaluate 
$$
\cos (-60^\circ)-\tan 135^\circ.
$$

- Since cosine is an even function, 
  $$
  \cos (-60^\circ)=\cos 60^\circ=\frac{1}{2}.
  $$
- The tangent of $$135^\circ$$ is given by
  $$
  \tan 135^\circ=\tan(180^\circ-45^\circ)=-\tan 45^\circ=-1.
  $$
- Therefore,
  $$
  \cos (-60^\circ)-\tan 135^\circ=\frac{1}{2}-(-1)=\frac{1}{2}+1=\frac{3}{2}.
  $$

---

2. Prove that
$$
\frac{\cos x}{1-\sin x}-\frac{\cos x}{1+\sin x}=2\tan x.
$$

- Write the difference over a common denominator:
  $$
  \frac{\cos x}{1-\sin x}-\frac{\cos x}{1+\sin x}=\frac{\cos x(1+\sin x)-\cos x(1-\sin x)}{(1-\sin x)(1+\sin x)}.
  $$
- Simplify the numerator:
  $$
  \cos x(1+\sin x)-\cos x(1-\sin x)=\cos x\big[(1+\sin x)-(1-\sin x)\big]=\cos x\,(2\sin x)=2\cos x\,\sin x.
  $$
- The denominator simplifies by the difference of squares:
  $$
  (1-\sin x)(1+\sin x)=1-\sin^2x.
  $$
- Recall the Pythagorean identity:
  $$
  1-\sin^2x=\cos^2x.
  $$
- Thus, the expression becomes:
  $$
  \frac{2\cos x\,\sin x}{\cos^2x}=\frac{2\sin x}{\cos x}=2\tan x.
  $$

---

3. Determine the general solution of 
$$
\cos^2 x-3\sin x-3=-3\sin^2 x.
$$

- First, bring all terms to one side:
  $$
  \cos^2 x-3\sin x-3+3\sin^2 x=0.
  $$
- Rewrite the expression by grouping:
  $$
  \cos^2 x+3\sin^2 x-3\sin x-3=0.
  $$
- Express $$\cos^2x$$ in terms of $$\sin x$$ using $$\cos^2x=1-\sin^2x$$:
  $$
  (1-\sin^2x)+3\sin^2x-3\sin x-3=0.
  $$
- Combine like terms:
  $$
  1+2\sin^2x-3\sin x-3=0\quad\Longrightarrow\quad 2\sin^2x-3\sin x-2=0.
  $$
- Let $$u=\sin x$$. The quadratic becomes:
  $$
  2u^2-3u-2=0.
  $$
- Compute the discriminant:
  $$
  \Delta = (-3)^2-4(2)(-2)=9+16=25.
  $$
- Solve for $$u$$:
  $$
  u=\frac{3\pm\sqrt{25}}{4}=\frac{3\pm5}{4}.
  $$
  This gives
  $$
  u_1=\frac{8}{4}=2 \quad \text{and} \quad u_2=\frac{-2}{4}=-\frac{1}{2}.
  $$
- Since $$\sin x$$ must lie between $$-1$$ and $$1$$, we discard $$u_1=2$$. Therefore, we have:
  $$
  \sin x=-\frac{1}{2}.
  $$
- The general solution for $$\sin x=-\frac{1}{2}$$ is:
  $$
  x=7\pi/6+2\pi k\quad \text{or} \quad x=11\pi/6+2\pi k,\quad k\in\mathbb{Z}.
  $$
1. Evaluate $$ \cos(-60^\circ) - \tan 135^\circ $$: - $$ \cos(-60^\circ) = \frac{1}{2} $$ - $$ \tan 135^\circ = -1 $$ - Result: $$ \frac{1}{2} - (-1) = \frac{3}{2} $$ 2. Prove that $$ \frac{\cos x}{1 - \sin x} - \frac{\cos x}{1 + \sin x} = 2\tan x $$: - Simplify the expression to $$ 2\tan x $$ using trigonometric identities. 3. Solve $$ \cos^2 x - 3\sin x - 3 = -3\sin^2 x $$: - General solution: $$ x = \frac{7\pi}{6} + 2\pi k $$ or $$ x = \frac{11\pi}{6} + 2\pi k $$, where $$ k $$ is an integer.

4.4 Prove that:

4.5 Determine the general solution of

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4.4 Prove that: 4.5 Determine the general solution of