1. Problem 5: A pole 6 m high casts a shadow 2.3 m long. Find the Sun's elevation angle. - The elevation angle $\theta$ satisfies $\tan \theta = \frac{\text{height}}{\text{shadow length}}$. - So, $\tan \theta = \frac{6}{2.3}$. - Calculate $\tan \theta \approx 2.6087$. - Find $\theta = \arctan(2.6087) \approx 69.1^\circ$. 2. Problem 6: Prove that $\sqrt{1 - \cos^2 \theta} \sec^2 \theta = \tan \theta$. - Use identity $\sin^2 \theta + \cos^2 \theta = 1 \Rightarrow \sin \theta = \sqrt{1-\cos^2 \theta}$. - Left side = $\sin \theta \sec^2 \theta = \sin \theta \frac{1}{\cos^2 \theta} = \frac{\sin \theta}{\cos^2 \theta}$. - Right side = $\tan \theta = \frac{\sin \theta}{\cos \theta}$. - Left side $\neq$ right side unless multiplied by $\frac{1}{\cos \theta}$; generally not equal. Thus original equality holds only under specific condition or is incorrect. 3. Problem 7: If $\cos A + \cos^2 A = 1$, prove $\sin^2 A + \sin^3 A = 1$. - From $\cos A + \cos^2 A = 1$. - Use $\sin^2 A = 1 - \cos^2 A$. - Express $\sin^3 A = \sin A \sin^2 A$. - Complexity suggests testing for specific $A$ or rewrite to verify equality. 4. Problem 8: Verify $(\tan \theta + 2)(2 \tan \theta + 1) = 5 \tan \theta + \sec^2 \theta$. - Expand left: $2 \tan^2 \theta + \tan \theta + 4 \tan \theta + 2 = 2 \tan^2 \theta + 5 \tan \theta + 2$. - Right: $5 \tan \theta + \sec^2 \theta$. - Use $\sec^2 \theta = 1 + \tan^2 \theta$. - Right becomes $5 \tan \theta + 1 + \tan^2 \theta$. - Compare: Left $2 \tan^2 \theta + 5 \tan \theta + 2$, Right $\tan^2 \theta + 5 \tan \theta + 1$. - Difference $\tan^2 \theta +1$, not equal. 5. Problem 9: Prove $\sin^6 \theta + \cos^6 \theta + 3 \sin^2 \theta \cos^2 \theta =1$. - Use sum of cubes: $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$ with $a=\sin^2 \theta$, $b=\cos^2 \theta$. - $\sin^6 \theta + \cos^6 \theta = (\sin^2 \theta + \cos^2 \theta)^3 - 3 \sin^2 \theta \cos^2 \theta (\sin^2 \theta + \cos^2 \theta) = 1 - 3 \sin^2 \theta \cos^2 \theta$. - Add $3 \sin^2 \theta \cos^2 \theta$: $1 - 3 \sin^2 \theta \cos^2 \theta + 3 \sin^2 \theta \cos^2 \theta = 1$. - Hence proved. 6. Problem 10: Prove $(\sin^4 \theta - \cos^4 \theta + 1) \csc^2 \theta = 2$. - Note $\sin^4 \theta - \cos^4 \theta = (\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta) = \sin^2 \theta - \cos^2 \theta$. - So expression $= (\sin^2 \theta - \cos^2 \theta + 1) \csc^2 \theta$. - Use $\csc^2 \theta = \frac{1}{\sin^2 \theta}$. - Expression $= \frac{\sin^2 \theta - \cos^2 \theta + 1}{\sin^2 \theta}$. - Rewrite numerator: $\sin^2 \theta + 1 - \cos^2 \theta$. - Using $\sin^2 \theta + \cos^2 \theta =1$, so numerator $= \sin^2 \theta + 1 - (1 - \sin^2 \theta) = \sin^2 \theta + 1 - 1 + \sin^2 \theta = 2 \sin^2 \theta$. - Expression $= \frac{2 \sin^2 \theta}{\sin^2 \theta} = 2$. - Proven. 7. Problem 11: If $\sin \theta + \cos \theta = \sqrt{3}$, prove $\tan \theta + \cot \theta = 1$. - Square left side: $(\sin \theta + \cos \theta)^2 = \sin^2 \theta + 2 \sin \theta \cos \theta + \cos^2 \theta = 3$. - Using $\sin^2 \theta + \cos^2 \theta = 1$, get $1 + 2 \sin \theta \cos \theta = 3$. - So, $2 \sin \theta \cos \theta = 2$, giving $\sin \theta \cos \theta = 1$. - $\tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{1} = 1$. - Proven. 8. Problem 12: Prove $\frac{\sin \theta}{1+\cos \theta} + \frac{1+\cos \theta}{\sin \theta} = 2 \csc \theta$. - Find common denominator $\sin \theta (1+\cos \theta)$. - Left side $= \frac{\sin^2 \theta + (1+\cos \theta)^2}{\sin \theta (1+\cos \theta)}$. - Expand numerator: $\sin^2 \theta + 1 + 2 \cos \theta + \cos^2 \theta$. - Using $\sin^2 \theta + \cos^2 \theta = 1$, numerator = $1 + 1 + 2 \cos \theta = 2(1 + \cos \theta)$. - Left side $= \frac{2(1 + \cos \theta)}{\sin \theta (1+\cos \theta)} = \frac{2}{\sin \theta} = 2 \csc \theta$. - Proven. 9. Problem 13: If $\tan A = \frac{3}{2}$, find $\sin A \cos A$. - $\tan A = \frac{\sin A}{\cos A} = \frac{3}{2}$. - Let $\sin A = 3k$, $\cos A = 2k$. - Use $\sin^2 A + \cos^2 A = 1$, thus $(3k)^2 + (2k)^2 = 1 \Rightarrow 9k^2 + 4k^2 = 1 \Rightarrow 13k^2 = 1 \Rightarrow k^2 = \frac{1}{13}$. - Then, $\sin A \cos A = 3k \times 2k = 6k^2 = \frac{6}{13}$. 10. Problem 14: Prove $(\sin x + \cos x)(\tan x + \cot x) = \sec x + \csc x$. - Note $\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x}$. - Left side $= (\sin x + \cos x) \frac{1}{\sin x \cos x} = \frac{\sin x + \cos x}{\sin x \cos x}$. - Right side $= \frac{1}{\cos x} + \frac{1}{\sin x} = \frac{\sin x + \cos x}{\sin x \cos x}$. - Both sides equal. Final answers: - Problem 5: Sun's elevation = $69.1^\circ$. - Problem 6: Equality does not hold generally. - Problem 7: Not directly proved, may require additional info. - Problem 8: Not equal generally. - Problem 9: Proved. - Problem 10: Proved. - Problem 11: Proved. - Problem 12: Proved. - Problem 13: $\sin A \cos A = \frac{6}{13}$. - Problem 14: Proved.