In \(\triangle OAC\), \(OA = OC\) (radii) \(\Rightarrow \angle OCA = \angle OAC\). \(\angle OCD = 90^\circ\) (radius \(\perp\) tangent). \(\Rightarrow \angle OCA + \angle ACD = 90^\circ\) \(\Rightarrow \angle OAC + \angle ACD = 90^\circ\) Hence \(\angle BAC + \angle ACD = 90^\circ\).

\(\triangle OAP \cong \triangle OAS \Rightarrow \angle 1 = \angle 2\). Similarly [by SSS Similarity criteria] \(\angle 3 = \angle 4,\ \angle 5 = \angle 6,\ \angle 7 = \angle 8\). Sum around O = \(360^\circ\): \(2(\angle 1+\angle 4+\angle 5+\angle 8)=360^\circ\) \(\Rightarrow \angle AOB + \angle COD = 180^\circ\). Similarly \(\angle BOC + \angle AOD = 180^\circ\).

To prove the identity \(\frac{\tan \theta}{1 – \cot \theta} + \frac{\cot \theta}{1 – \tan \theta} = 1 + \sec \theta \csc \theta\), we will work with the Left Hand Side (LHS) and simplify it to match the Right Hand Side (RHS).
Step 1: Express in terms of \(\sin \theta\) and \(\cos \theta\) Substitute \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and \(\cot \theta = \frac{\cos \theta}{\sin \theta}\) into the expression: \[LHS = \frac{\frac{\sin \theta}{\cos \theta}}{1 – \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 – \frac{\sin \theta}{\cos \theta}}\]
Step 2: Simplify the denominators Find a common denominator for the terms within the denominators: \[LHS = \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta – \cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta – \sin \theta}{\cos \theta}}\] Step 3: Multiply by the reciprocals Simplify the complex fractions: \[LHS = \frac{\sin^2 \theta}{\cos \theta (\sin \theta – \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta (\cos \theta – \sin \theta)}\] To have a common denominator, rewrite \((\cos \theta – \sin \theta)\) as \(-(\sin \theta – \cos \theta)\): \[LHS = \frac{\sin^2 \theta}{\cos \theta (\sin \theta – \cos \theta)} – \frac{\cos^2 \theta}{\sin \theta (\sin \theta – \cos \theta)}\]
Step 4: Combine the fractions Now, find the common denominator, which is \(\sin \theta \cos \theta (\sin \theta – \cos \theta)\): \[LHS = \frac{\sin^3 \theta – \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta – \cos \theta)}\]
Step 5: Factor the numerator Use the algebraic identity \(a^3 – b^3 = (a – b)(a^2 + ab + b^2)\): \[LHS = \frac{(\sin \theta – \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta \cos \theta (\sin \theta – \cos \theta)}\] Cancel the common factor \((\sin \theta – \cos \theta)\) from the numerator and denominator: \[LHS = \frac{\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta}{\sin \theta \cos \theta}\]
Step 6: Use Pythagorean identity and simplify Apply \(\sin^2 \theta + \cos^2 \theta = 1\): \[LHS = \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta}\] Separate the terms: \[LHS = \frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta}\] \[LHS = \left(\frac{1}{\cos \theta}\right) \left(\frac{1}{\sin \theta}\right) + 1\] Since \(\frac{1}{\cos \theta} = \sec \theta\) and \(\frac{1}{\sin \theta} = \csc \theta\): \[LHS = \sec \theta \csc \theta + 1\]
Answer: \(1 + \sec \theta \csc \theta = RHS\)

Prove that: $$\frac{\sin A + \cos A}{\sin A – \cos A} + \frac{\sin A – \cos A}{\sin A + \cos A} = \frac{2}{2 \sin^2 A – 1}$$ Proof:
Step 1: Combine the fractions using a common denominator. $$LHS = \frac{(\sin A + \cos A)^2 + (\sin A – \cos A)^2}{(\sin A – \cos A)(\sin A + \cos A)}$$
Step 2: Expand the numerator and simplify the denominator using the difference of squares. $$= \frac{(\sin^2 A + 2\sin A\cos A + \cos^2 A) + (\sin^2 A – 2\sin A\cos A + \cos^2 A)}{\sin^2 A – \cos^2 A}$$
Step 3: Cancel terms and apply the identity \(\sin^2 A + \cos^2 A = 1\). $$LHS = \frac{2(\sin^2 A + \cos^2 A)}{\sin^2 A – \cos^2 A} = \frac{2}{\sin^2 A – \cos^2 A}$$
Step 4: Substitute \(\cos^2 A = 1 – \sin^2 A\) to match the target denominator. $$LHS = \frac{2}{\sin^2 A – (1 – \sin^2 A)} = \frac{2}{2\sin^2 A – 1}$$ Therefore, \(LHS = RHS\).

Let ratio \(k:1\), point on y-axis \((0,y)\). \(0 = \frac{-k+5}{k+1} \Rightarrow k=5\). \(y = \frac{-4(5) + (-6)(1)}{5+1}\) \(= \frac{-20-6}{6} = \frac{-26}{6} = -\frac{13}{3}\).

Let radius = r, height of cylinder = h. Given \(r = \frac{h}{2} \Rightarrow h = 2r\). Volume = \(\frac{2}{3}\pi r^3 + \pi r^2 h\) \(= \frac{2}{3}\pi r^3 + \pi r^2(2r)\) \(= \pi r^3(\frac{2}{3}+2)\) \(= \pi r^3 \times \frac{8}{3}\) \(\frac{1408}{21} = \frac{8}{3} \times \frac{22}{7} \times r^3 \) ⇒ \(r^3 = 8 \Rightarrow r=2\) m, h = 4 m.

Total outcomes = 36. Favorable outcomes (difference 2): (1,3),(3,1),(2,4),(4,2),(3,5),(5,3),(4,6),(6,4) → 8 outcomes. \(P = \frac{8}{36} = \frac{2}{9}\).