Detomine the value of the following expression:- \[ \frac{\tan \left(180^{\circ}-x\right) \cdot \sin \left(x-90^{\circ}\right)}{4 \sin \left(360^{\circ}+x\right)} \] If \( \sin A=p \) and \( \cos A=q \) - Write \( \tan A \) in items of pand \( q \) Simplify \( p^{4}-q^{4} \) to a single Simplify trigonometric ratio 1 Prove that:- \( \frac{\cos \theta}{\sin \theta}-\frac{\cos 2 \theta}{\sin \theta \cdot \cos \theta}(5) \) Determine the general Solution of:- \( { }^{9} \) (5) \[ 2 \operatorname{Sin} 2 x+3 \operatorname{Sin} x=0(6) \]

1. Evaluate \[ \frac{\tan (180^\circ-x)\cdot\sin (x-90^\circ)}{4\sin (360^\circ+x)}. \] **Step 1.1. Simplify \(\tan(180^\circ-x)\):** Recall that \[ \tan(180^\circ-x)= -\tan x. \] **Step 1.2. Simplify \(\sin(x-90^\circ)\):** Using the sine subtraction formula or standard identities, note that \[ \sin(x-90^\circ) = -\cos x. \] **Step 1.3. Simplify \(\sin(360^\circ+x)\):** Since sine is periodic with period \(360^\circ\), we have \[ \sin(360^\circ+x)= \sin x. \] **Step 1.4. Substitute the simplified parts:** The expression becomes \[ \frac{(-\tan x)(-\cos x)}{4\sin x} = \frac{\tan x\,\cos x}{4 \sin x}. \] **Step 1.5. Express \(\tan x\) in terms of \(\sin x\) and \(\cos x\):** Remember that \[ \tan x=\frac{\sin x}{\cos x}. \] Thus, \[ \frac{\frac{\sin x}{\cos x}\cdot\cos x}{4\sin x}= \frac{\sin x}{4\sin x}= \frac{1}{4}, \] provided \(\sin x\ne0\). The value of the expression is \[ \frac{1}{4}. \] --- 2. Given \(\sin A=p\) and \(\cos A=q\), write \(\tan A\) in terms of \(p\) and \(q\). Since \[ \tan A = \frac{\sin A}{\cos A}, \] we have \[ \tan A= \frac{p}{q}. \] --- 3. Simplify \( p^4 - q^4 \) to a single expression. **Step 3.1. Factor the expression:** Recall that \[ a^4-b^4=(a^2-b^2)(a^2+b^2). \] Thus, \[ p^4-q^4= (p^2-q^2)(p^2+q^2). \] **Step 3.2. Use the Pythagorean identity:** Since \(p=\sin A\) and \(q=\cos A\), we have \[ p^2+q^2= \sin^2A+\cos^2A=1. \] Therefore, \[ p^4-q^4= p^2-q^2. \] --- 4. Prove that \[ \frac{\cos\theta}{\sin\theta}-\frac{\cos2\theta}{\sin\theta\cdot\cos\theta} = \tan\theta. \] **Step 4.1. Write the expression with a common denominator:** \[ \frac{\cos\theta}{\sin\theta}-\frac{\cos2\theta}{\sin\theta\cos\theta}= \frac{\cos\theta\cdot\cos\theta-\cos2\theta}{\sin\theta\cos\theta} = \frac{\cos^2\theta-\cos2\theta}{\sin\theta\cos\theta}. \] **Step 4.2. Use the double-angle identity:** Recall that \[ \cos2\theta=2\cos^2\theta-1. \] Thus, \[ \cos^2\theta-\cos2\theta= \cos^2\theta- (2\cos^2\theta-1)= 1-\cos^2\theta. \] **Step 4.3. Replace using the Pythagorean identity:** Since \[ 1-\cos^2\theta= \sin^2\theta, \] the expression becomes \[ \frac{\sin^2\theta}{\sin\theta\cos\theta}= \frac{\sin\theta}{\cos\theta}= \tan\theta. \] --- 5. Determine the general solution of \[ 2\sin2x+3\sin x=0. \] **Step 5.1. Express \(\sin2x\) in terms of \(\sin x\) and \(\cos x\):** Recall that \[ \sin2x= 2\sin x\cos x. \] Substitute into the equation: \[ 2(2\sin x\cos x)+3\sin x= 4\sin x\cos x+3\sin x= 0. \] **Step 5.2. Factor out \(\sin x\):** \[ \sin x\,(4\cos x+3)= 0. \] **Step 5.3. Set each factor equal to zero:** (i) \(\sin x= 0;\) The general solution is \[ x= n\pi,\quad n\in \mathbb{Z}. \] (ii) \(4\cos x+3= 0;\) Solve for \(\cos x\): \[ \cos x= -\frac{3}{4}. \] The general solution for \(\cos x = -\frac{3}{4}\) is \[ x= \arccos\Bigl(-\frac{3}{4}\Bigr) + 2\pi k\quad \text{or}\quad x= -\arccos\Bigl(-\frac{3}{4}\Bigr) + 2\pi k,\quad k\in\mathbb{Z}. \] --- Each part has been simplified and solved as indicated above.