1. If \(x^2 = \sin^2 30^\circ + 4\cot^2 45^\circ - \sec^2 60^\circ\), then the value of \(x\) is —
(a) \(\pm 1\) (b) \(\pm \cfrac{1}{2}\) (c) \(\pm \cfrac{1}{\sqrt2}\) (d) \(\pm \cfrac{1}{\sqrt3}\)
2. If \(x^2 = \sin^2 30^\circ + 4\cot^2 45^\circ - \sec^2 60^\circ\), then what is the value of \(x\)?
3. If \(x^2 = \sin^2 30^\circ + 4\cot^2 45^\circ - \sec^2 60^\circ\), find the value of \(x\). Let me know if you'd like a step-by-step solution next. I'm ready when you are.
4. If \(x^2 = \sin^2 30^\circ + 4\cot^2 45^\circ - \sec^2 60^\circ\), find the value of \(x\).
5. If \(x^2 = \sin^2 30^\circ + 4\cot^2 45^\circ - \sec^2 60^\circ\), then find the value of \(x\).
6. If \(x^2 = \sin^2 30^\circ + 4\cot^2 45^\circ - \sec^2 60^\circ\), then find the value of \(x\).
7. Show that \[ \frac{2 \tan^2 30^\circ}{1 - \tan^2 30^\circ} + \sec^2 45^\circ - \cot^2 45^\circ = \sec 60^\circ \]
8. Evaluate the expression: \[ \cot^2 30^\circ - 2\cos^2 60^\circ - 4\sin^2 30^\circ - \frac{3}{4}\sec^2 45^\circ + \tan 45^\circ \]
9. Show that: \[ \cfrac{2 \tan^2 30^\circ}{1 - \tan^2 30^\circ} + \sec^2 45^\circ - \cot^2 45^\circ = \sec 60^\circ \]
10. Evaluate: \[ 3\tan^2 45^\circ - \sin^2 60^\circ - \frac{1}{3} \cot^2 30^\circ - \frac{1}{8} \sec^2 45^\circ \]
11. If \(x = \sin^2 30^\circ + 4 \cot^2 45^\circ - \sec^2 60^\circ\), find the value of \(x\).
12. Evaluate the expression: \[ \frac{1 - \sin^2 30^\circ}{1 + \sin^2 45^\circ} \times \frac{\cos^2 60^\circ + \cos^2 30^\circ}{\csc^2 90^\circ - \cot^2 90^\circ} \div (\sin 60^\circ \tan 30^\circ) \]
13. Find the value of: \[ \frac{2\tan^2 30^\circ}{1 - \tan^2 30^\circ} + \sec^2 45^\circ - \cot^2 45^\circ - \sec 60^\circ \]
14. If \[ x \cos 60^\circ = \frac{2 \tan 45^\circ}{1 + \tan^2 45^\circ} - \frac{1 - \tan^2 30^\circ}{1 + \tan^2 30^\circ} \] then find the value of \(x\).
15. Evaluate the expression: \[ \frac{4}{3} \cot^2 30^\circ + 3 \sin^2 60^\circ - 2 \csc^2 60^\circ - \frac{3}{4} \tan^2 30^\circ \]
16. 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)
17. Find the value of \(x\) from the equation: \(x \sec^2 45^\circ \cdot \csc^2 45^\circ + 2(\sin 60^\circ + \sin 30^\circ) = \tan 60^\circ\)
18. Evaluate the expression: \[ \frac{(\sin 0^\circ + \sin 60^\circ)(\cos 60^\circ + \cot 45^\circ)}{(\cot 60^\circ + \tan 30^\circ)(\csc 30^\circ - \csc 90^\circ)} \]
19. The English translation is: "If \(x^2 = \sin^2 30° + 4\cot^2 45° - \sec^2 60°\), determine the value of \(x\)."
20. Show that: \[ \sqrt{\cfrac{1 + \cos 30^\circ}{1 - \cos 30^\circ}} = \sec 60^\circ + \tan 60^\circ \]
21. Determine the value of \[ \sec^2 60^\circ - \cot^2 30^\circ - \frac{2 \tan 30^\circ \cdot \csc 60^\circ}{1 + \tan^2 30^\circ} \]
22. \[ \tan 60^\circ = \sqrt{3},\quad \tan 30^\circ = \frac{1}{\sqrt{3}} \\ \cos 60^\circ = \frac{1}{2},\quad \cos 30^\circ = \frac{\sqrt{3}}{2} \\ \sin 60^\circ = \frac{\sqrt{3}}{2},\quad \sin 30^\circ = \frac{1}{2} \] \[ \frac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \cdot \tan 30^\circ} = \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + \sqrt{3} \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{3 - 1}{\sqrt{3}}}{2} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} \] \[ \cos 60^\circ \cos 30^\circ + \sin 60^\circ \sin 30^\circ = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \] \[ \frac{1}{\sqrt{3}} + \frac{\sqrt{3}}{2} = \frac{2 + 3}{2\sqrt{3}} = \frac{5}{2\sqrt{3}} \]
23. Evaluate the following expression: \[ \sec^2 60^\circ - \cot^2 30^\circ - \frac{2 \tan 30^\circ \csc 60^\circ}{1 + \tan^2 30^\circ} \]
24. Determine the value of \[ \sec^2 60^\circ - \cot^2 30^\circ \cdot \cfrac{2 \tan 60^\circ \cdot \csc 60^\circ}{1 + \tan^2 30^\circ} \]
