1. Stating the problem: Solve the trigonometric equations $$\sqrt{\frac{1 + \sin \Delta}{1 - \sin \Delta}} = \tan \left(\frac{\pi}{4} + \frac{\Delta}{2}\right)$$ and $$\sin 2\Delta + \sin 5\Delta - \sin \Delta = \tan^2 \Delta$$ 2. For the first equation, recall the identity: $$\tan\left(\frac{\pi}{4} + x\right) = \frac{1 + \tan x}{1 - \tan x}$$ and the half-angle formulas: $$\tan\left(\frac{\pi}{4} + \frac{\Delta}{2}\right) = \frac{1 + \tan(\frac{\Delta}{2})}{1 - \tan(\frac{\Delta}{2})}$$ Also, the expression under the square root can be rewritten using the identity: $$\sqrt{\frac{1 + \sin \Delta}{1 - \sin \Delta}} = \frac{1 + \sin \Delta}{\cos \Delta}$$ because $$\frac{1 + \sin \Delta}{1 - \sin \Delta} = \left(\frac{1 + \sin \Delta}{\cos \Delta}\right)^2$$ 3. Using the half-angle substitution: $$\tan\left(\frac{\pi}{4} + \frac{\Delta}{2}\right) = \frac{1 + \tan(\frac{\Delta}{2})}{1 - \tan(\frac{\Delta}{2})}$$ and $$\frac{1 + \sin \Delta}{\cos \Delta} = \frac{1 + 2 \sin(\frac{\Delta}{2}) \cos(\frac{\Delta}{2})}{\cos \Delta}$$ 4. Simplify and verify equality: Since both sides simplify to the same expression, the equation holds true for all $\Delta$ where defined. 5. For the second equation: $$\sin 2\Delta + \sin 5\Delta - \sin \Delta = \tan^2 \Delta$$ Use sum-to-product formulas: $$\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$$ Apply to $\sin 2\Delta + \sin 5\Delta$: $$2 \sin \frac{7\Delta}{2} \cos \frac{-3\Delta}{2} = 2 \sin \frac{7\Delta}{2} \cos \frac{3\Delta}{2}$$ So the left side becomes: $$2 \sin \frac{7\Delta}{2} \cos \frac{3\Delta}{2} - \sin \Delta$$ 6. The equation is: $$2 \sin \frac{7\Delta}{2} \cos \frac{3\Delta}{2} - \sin \Delta = \tan^2 \Delta$$ 7. This is a transcendental equation; solutions can be found numerically or by testing special angles. 8. For example, test $\Delta = 0$: Left: $0 + 0 - 0 = 0$ Right: $\tan^2 0 = 0$ So $\Delta=0$ is a solution. 9. Test $\Delta = \frac{\pi}{4}$: Left: $\sin \frac{\pi}{2} + \sin \frac{5\pi}{4} - \sin \frac{\pi}{4} = 1 + (-\frac{\sqrt{2}}{2}) - \frac{\sqrt{2}}{2} = 1 - \sqrt{2} \approx -0.414$ Right: $\tan^2 \frac{\pi}{4} = 1$ Not equal, so $\Delta=\frac{\pi}{4}$ is not a solution. 10. Final answers: - Equation 2 holds for all $\Delta$ where defined. - Equation 3 has solution $\Delta=0$ and possibly others found numerically.