Exam-Level Practice: Advanced Trigonometric Proofs

Practice & Review

Test your skills by proving challenging trigonometric identities. These problems require combining multiple strategies, such as converting to sin/cos, using conjugates, factoring, and applying Pythagorean identities. Success here demonstrates a deep mastery of the topic.

Exam-Level Practice: Advanced Trigonometric Proofs

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Prove the identity: $$ \frac{\sec(x) - \cos(x)}{\sec(x)} = \sin^2(x) $$

Work with the left side (LHS). $$ \text{LHS} = \frac{\frac{1}{\cos(x)} - \cos(x)}{\frac{1}{\cos(x)}} $$ Multiply numerator and denominator by $\cos(x)$: $$ = \left( \frac{1}{\cos(x)} - \cos(x) \right) \cos(x) = 1 - \cos^2(x) = \sin^2(x) $$ LHS = RHS. Q.E.D.

Prove the identity: $$ \frac{\sin(x)}{1 - \cos(x)} = \csc(x) + \cot(x) $$

Work with the LHS. Multiply by the conjugate $1+\cos(x)$: $$ \text{LHS} = \frac{\sin(x)}{1 - \cos(x)} \cdot \frac{1+\cos(x)}{1+\cos(x)} = \frac{\sin(x)(1+\cos(x))}{1-\cos^2(x)} $$ $$ = \frac{\sin(x)(1+\cos(x))}{\sin^2(x)} = \frac{1+\cos(x)}{\sin(x)} = \frac{1}{\sin(x)} + \frac{\cos(x)}{\sin(x)} = \csc(x) + \cot(x) $$ LHS = RHS. Q.E.D.

Prove: $$ (\sec(\theta) + \tan(\theta))(1 - \sin(\theta)) = \cos(\theta) $$

Work with the LHS. Convert to $\sin$ and $\cos$: $$ \text{LHS} = \left( \frac{1}{\cos(\theta)} + \frac{\sin(\theta)}{\cos(\theta)} \right) (1-\sin(\theta)) = \left( \frac{1+\sin(\theta)}{\cos(\theta)} \right) (1-\sin(\theta)) $$ $$ = \frac{(1+\sin(\theta))(1-\sin(\theta))}{\cos(\theta)} = \frac{1-\sin^2(\theta)}{\cos(\theta)} = \frac{\cos^2(\theta)}{\cos(\theta)} = \cos(\theta) $$ LHS = RHS. Q.E.D.

Prove the identity: $$ \frac{\cot^2(t)}{1 + \csc(t)} = \frac{1 - \sin(t)}{\sin(t)} $$

Work with the LHS. Use Pythagorean identity $\cot^2(t) = \csc^2(t) - 1$: $$ \text{LHS} = \frac{\csc^2(t)-1}{1+\csc(t)} = \frac{(\csc(t)-1)(\csc(t)+1)}{1+\csc(t)} $$ $$ = \csc(t)-1 = \frac{1}{\sin(t)}-1 = \frac{1-\sin(t)}{\sin(t)} $$ LHS = RHS. Q.E.D.

Prove the identity: $$ \tan^2(x) - \sin^2(x) = \tan^2(x)\sin^2(x) $$

Work with the LHS. Convert $\tan^2(x)$ to $\sin^2(x)/\cos^2(x)$: $$ \text{LHS} = \frac{\sin^2(x)}{\cos^2(x)} - \sin^2(x) = \sin^2(x) \left( \frac{1}{\cos^2(x)} - 1 \right) $$ $$ = \sin^2(x) (\sec^2(x)-1) = \sin^2(x)\tan^2(x) $$ LHS = RHS. Q.E.D.

Prove: $$ \frac{\cos(x)}{1 - \tan(x)} + \frac{\sin(x)}{1 - \cot(x)} = \sin(x) + \cos(x) $$

Convert to $\sin$ and $\cos$ on LHS: $$ \frac{\cos(x)}{1 - \frac{\sin(x)}{\cos(x)}} + \frac{\sin(x)}{1 - \frac{\cos(x)}{\sin(x)}} = \frac{\cos^2(x)}{\cos(x) - \sin(x)} + \frac{\sin^2(x)}{\sin(x) - \cos(x)} $$ $$ = \frac{\cos^2(x)}{\cos(x) - \sin(x)} - \frac{\sin^2(x)}{\cos(x) - \sin(x)} = \frac{\cos^2(x)-\sin^2(x)}{\cos(x)-\sin(x)} $$ $$ = \frac{(\cos(x)-\sin(x))(\cos(x)+\sin(x))}{\cos(x)-\sin(x)} = \cos(x)+\sin(x) $$ LHS = RHS. Q.E.D.

Prove: $$ \sqrt{\frac{1+\sin(x)}{1-\sin(x)}} = |\sec(x) + \tan(x)| $$

Work with LHS. Multiply inside the square root by the conjugate: $$ \text{LHS} = \sqrt{\frac{1+\sin(x)}{1-\sin(x)} \cdot \frac{1+\sin(x)}{1+\sin(x)}} = \sqrt{\frac{(1+\sin(x))^2}{1-\sin^2(x)}} $$ $$ = \sqrt{\frac{(1+\sin(x))^2}{\cos^2(x)}} = \frac{\sqrt{(1+\sin(x))^2}}{\sqrt{\cos^2(x)}} = \frac{|1+\sin(x)|}{|\cos(x)|} $$ Since $1+\sin(x) \ge 0$, this is $ \frac{1+\sin(x)}{|\cos(x)|} = |\frac{1}{\cos(x)} + \frac{\sin(x)}{\cos(x)}| = |\sec(x) + \tan(x)| $. Q.E.D.

Prove: $$ (\sin\alpha + \cos\alpha)(\tan\alpha + \cot\alpha) = \sec\alpha + \csc\alpha $$

LHS: Convert tan and cot to sin/cos. $$ (\sin\alpha + \cos\alpha)(\frac{\sin\alpha}{\cos\alpha} + \frac{\cos\alpha}{\sin\alpha}) = (\sin\alpha + \cos\alpha)(\frac{\sin^2\alpha + \cos^2\alpha}{\sin\alpha\cos\alpha}) $$ $$ = (\sin\alpha + \cos\alpha)(\frac{1}{\sin\alpha\cos\alpha}) = \frac{\sin\alpha}{\sin\alpha\cos\alpha} + \frac{\cos\alpha}{\sin\alpha\cos\alpha} $$ $$ = \frac{1}{\cos\alpha} + \frac{1}{\sin\alpha} = \sec\alpha + \csc\alpha $$ LHS = RHS. Q.E.D.

Prove the identity: $$ \sec^4(x) - \sec^2(x) = \tan^4(x) + \tan^2(x) $$

Work with the LHS. Factor out $\sec^2(x)$. $$ \text{LHS} = \sec^2(x)(\sec^2(x) - 1) $$ Use Pythagorean identities $\sec^2(x) = 1+\tan^2(x)$ and $\sec^2(x)-1 = \tan^2(x)$. $$ = (1+\tan^2(x))(\tan^2(x)) = \tan^2(x) + \tan^4(x) $$ LHS = RHS. Q.E.D.

Prove the identity: $$ \frac{1+\tan^2\theta}{1+\cot^2\theta} = \tan^2\theta $$

LHS: Use Pythagorean identities $1+\tan^2\theta = \sec^2\theta$ and $1+\cot^2\theta = \csc^2\theta$. $$ \text{LHS} = \frac{\sec^2\theta}{\csc^2\theta} = \frac{1/\cos^2\theta}{1/\sin^2\theta} $$ $$ = \frac{1}{\cos^2\theta} \cdot \frac{\sin^2\theta}{1} = \frac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta $$ LHS = RHS. Q.E.D.

Exam-Level Practice: Advanced Trigonometric Proofs

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