1. Solve: \[ \frac{1}{a + b + x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x} \] where \(x \ne 0\) and \(x \ne -(a + b)\).

2. Solve: \[ \frac{1}{a + b + x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}, \quad x \ne 0,\ -(a + b) \]

3. Solve: \[ \frac{1}{x} - \frac{1}{x + b} = \frac{1}{a} - \frac{1}{a + b}, \quad x \ne 0,\ -b \]

4. Solve: \[ \frac{1}{2a + b + 2x} = \frac{1}{2a} + \frac{1}{b} + \frac{1}{2x} \]

5. If \((b + c - a)x = (c + a - b)y = (a + b - c)z = 2\), then show that \[ \left(\frac{1}{x} + \frac{1}{y}\right)\left(\frac{1}{y} + \frac{1}{z}\right)\left(\frac{1}{z} + \frac{1}{x}\right) = abc \]

6. Solve: \[ \frac{1}{x} - \frac{1}{3} = \frac{1}{x + 2} - \frac{1}{5} \]

7. If \[ \frac{3 - 5x}{x} + \frac{3 - 5y}{y} + \frac{3 - 5z}{z} = 0 \] then what is the value of \[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z}? \]

8. If \(\frac{\csc θ + \sin θ}{\csc θ - \sin θ} = \frac{5}{2}\), Let \(\sin θ = x\), then \(\csc θ = \frac{1}{x}\) \[ \frac{\frac{1}{x} + x}{\frac{1}{x} - x} = \frac{5}{2} \Rightarrow \frac{1 + x^2}{1 - x^2} = \frac{5}{2} \Rightarrow 2(1 + x^2) = 5(1 - x^2) \Rightarrow 2 + 2x^2 = 5 - 5x^2 \Rightarrow 7x^2 = 3 \Rightarrow x = \sqrt{\frac{3}{7}} \] ∴ \(\sin θ = \sqrt{\frac{3}{7}}\)

9. If \(x = \sqrt{7} + \sqrt{6}\), then find the simplified value of \[ x - \frac{1}{x} \]

10. If \(x = \sqrt{7} + \sqrt{6}\), then find the simplified value of \[ x + \frac{1}{x} \]

11. If \(x = 2 + \sqrt{3}\) and \(y = 2 - \sqrt{3}\), then find the value of \[ x - \frac{1}{x} \]

12. If \(x = 2 + \sqrt{3}\), then find the value of \[ x + \frac{1}{x} \]

(a) 2 (b) 2√3 (c) 4 (d) 2-√3

13. If \(x = \sqrt{3} + \sqrt{2}\) and \(y = \frac{1}{x}\), then find the value of: \[ (x + \frac{1}{x})^2 + \left( \frac{1}{y} - y \right)^2 \]

14. Translate: Solve: \[ x + \frac{1}{x} = p + \frac{1}{p} \]

15. If \[ (b + c - a) x = (c + a - b) y = (a + b - c) z = 2 \] then prove that \[ \left( \frac{1}{y} + \frac{1}{z} \right) \left( \frac{1}{z} + \frac{1}{x} \right) \left( \frac{1}{x} + \frac{1}{y} \right) = abc \]

16. If \(\frac{a^2}{b + c} = \frac{b^2}{c + a} = \frac{c^2}{a + b} = 1\), then prove that \[\frac{1}{1 + a} + \frac{1}{1 + b} + \frac{1}{1 + c} = 1\]

17. If \[ \frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{a + b} \] then prove that each of these ratios is either \(\frac{1}{2}\) or \(-1\).

18. If \[ \frac{b + c - a}{y + z - x} = \frac{c + a - b}{z + x - y} = \frac{a + b - c}{x + y - z} \] then prove that \[ \frac{a}{x} = \frac{b}{y} = \frac{c}{z} \]

19. If \[ \frac{by + cz}{b^2 + c^2} = \frac{cz + ax}{c^2 + a^2} = \frac{ax + by}{a^2 + b^2} \] then prove that \[ \frac{x}{a} = \frac{y}{b} = \frac{z}{c} \]

20. If \(\angle A + \angle B = 90^\circ\), then prove that \[ 1 + \cfrac{\tan A}{\tan B} = \tan^2 A \sec^2 B \]

21. Left-hand side (LHS): \[ 1 + \cfrac{\tan A}{\tan B} = 1 + \cfrac{\tan(90^\circ - B)}{\tan B} = 1 + \cfrac{\cot B}{\cot B} = 1 + \cot^2 B = \csc^2 B \] Right-hand side (RHS): \[ \tan^2 A \cdot \sec^2 B = \tan^2(90^\circ - B) \cdot \sec^2 B = \cot^2 B \cdot \sec^2 B = \cfrac{\cos^2 B}{\sin^2 B} \cdot \cfrac{1}{\cos^2 B} = \cfrac{1}{\sin^2 B} = \csc^2 B \] \(\therefore\) LHS = RHS (Proved)

22. In triangle \(\triangle ABC\), prove that: \[ \sin\left(\frac{A + B}{2}\right) + \cos\left(\frac{B + C}{2}\right) = \cos\left(\frac{C}{2}\right) + \sin\left(\frac{A}{2}\right) \]

23. If \[ \frac{x^2 - yz}{a} = \frac{y^2 - zx}{b} = \frac{z^2 - xy}{c} \] then prove that \[ (a + b + c)(x + y + z) = ax + by + cz \]

24. Given: \[ \frac{x}{b + c} = \frac{y}{c + a} = \frac{z}{a + b} \] Prove that: \[ \frac{a}{y + z - x} = \frac{b}{z + x - y} = \frac{c}{x + y - z} \]

25. Given that \(A + B = 90^\circ\), prove that \[ \tan A + \tan B = \frac{\csc^2 B}{\sqrt{\csc^2 B - 1}} \]

26. Given: \[ x^2 : (by + cz) = y^2 : (cz + ax) = z^2 : (ax + by) = 1 \] Show that: \[ \frac{a}{a + x} + \frac{b}{b + y} + \frac{c}{c + z} = 1 \]

27. Solve: \[ \frac{1}{x - \frac{1}{p}} + \frac{1}{x - \frac{1}{q}} = p + q \] where \(x \ne \frac{1}{p}, \frac{1}{q}\)

28. If \[ \frac{a}{y + z} = \frac{b}{z + x} = \frac{c}{x + y} \] then prove that \[ \frac{a(b - c)}{y^2 - z^2} = \frac{b(c - a)}{z^2 - x^2} = \frac{c(a - b)}{x^2 - y^2} \]

29. If \(a + b + c = 0\), then prove that \[ \frac{bc}{2a^2 + bc} + \frac{ca}{2b^2 + ca} + \frac{ab}{2c^2 + ab} = 1 \]

30. Solve: \[ \cfrac{1}{x} - \cfrac{1}{x + b} = \cfrac{1}{a} - \cfrac{1}{a + b}, \quad x \ne 0, -b \]