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Trigonometry practice for beginners

The following problems need only trigonometric definition of functions and the equation

\[\sin^2{\theta} + \cos^2{\theta} = 1\]

unless otherwise stated. More will be written.

Prove the following:

  1. \(\csc{\theta} \cdot \cos{\theta} = \cot{\theta}\)

  2. \(\csc{\theta}\cdot\tan{\theta}=\sec{\theta}\)

  3. \(1+\tan^2(-\theta) = \sec^2{\theta}\)

  4. \(\cos{\theta}(\tan{\theta}+\cot{\theta})=\csc{\theta}\)

  5. \(\sin{\theta}(\tan{\theta}+\cot{\theta})=\sec{\theta}\)

  6. \(\tan{\theta}\cdot\cot{\theta}-\cos^2{\theta}=\sin^2{\theta}\)

  7. \(\sin{\theta}\cdot\csc{\theta}-\cos^2{\theta}=\sin^2{\theta}\)

  8. \((\sec{\theta}-1)(\sec{\theta}+1)=\tan^2{\theta}\)

  9. \((\csc{\theta}-1)(\csc{\theta}+1)=\cot^2{\theta}\)

  10. \((\sec{\theta}-\tan{\theta})(\sec{\theta}+\tan{\theta})=1\)

  11. \(\sin^2{\theta}(1+\cot^2{\theta})=1\)

  12. \((1-\sin^2{\theta})(1+\tan{\theta})=1\)

  13. \((\sin{\theta}+\cos{\theta})^2+(\sin{\theta}-\cos{\theta})=2\)

  14. \(\tan^2{\theta}\cdot\cos^2{\theta}+\cot^2{\theta}\cdot\sin^2{\theta}=1\)

  15. \(\sec^4{\theta}-\sec^2{\theta}=\tan^4{\theta}+\tan^2{\theta}\)

  16. \(\sec{\theta}-\tan{\theta}=\dfrac{\cos{\theta}}{1+\sin{\theta}}\)

  17. \(\csc{\theta}-\cot{\theta}=\dfrac{\sin{\theta}}{1+\sin{\theta}}\)

  18. \(3\sin^2{\theta}+4\cos^2{\theta}=3+\cos^2{\theta}\)

  19. \(9\sec^2{\theta}-5\tan^2{\theta}=5+4\sec^2{\theta}\)

  20. \(1-\dfrac{\cos^2{\theta}}{1+\sin{\theta}}=\sin{\theta}\)

  21. \(1-\dfrac{\sin^2{\theta}}{1-\cos{\theta}}=-\cos{\theta}\)

  22. \(\dfrac{1+\tan{\theta}}{1-\tan{\theta}}=\dfrac{\cot{\theta}+1}{\cot{\theta}-1}\)

  23. \(\dfrac{\sec{\theta}}{\csc{\theta}}+\dfrac{\sin{\theta}}{\cos{\theta}}=2\tan{\theta}\)

  24. \(\dfrac{\csc{\theta}-1}{\cot{\theta}}=\dfrac{\cot{\theta}}{\csc{\theta}+1}\)

  25. \(\dfrac{1+\sin{\theta}}{1-\sin{\theta}}=\dfrac{\csc{\theta}+1}{\csc{\theta}-1}\)

  26. \(\dfrac{1-\sin{\theta}}{\cos{\theta}}+\dfrac{\cos{\theta}}{1-\sin{\theta}}=2\sec{\theta}\)

  27. \(\dfrac{\sin{\theta}}{\sin{\theta}-\cos{\theta}}=\dfrac{1}{1-\cot{\theta}}\)

  28. \(1-\dfrac{\sin^2{\theta}}{1+\cos{\theta}}=\cos{\theta}\)

  29. \(\dfrac{1-\sin{\theta}}{1+\sin{\theta}}=(\sec{\theta}-\tan{\theta})^2\)

  30. \(\dfrac{\cos{\theta}}{1-\tan{\theta}}+\dfrac{\sin{\theta}}{1-\cot{\theta}}=\sin{\theta}+\cos{\theta}\)

  31. \(\dfrac{\cot{\theta}}{1-\tan{\theta}}+\dfrac{\tan{\theta}}{1-\cot{\theta}}=1+\tan{\theta}+\cot{\theta}\)

  32. \(\tan{\theta}+\dfrac{\cos{\theta}}{1+\sin{\theta}}=\sec{\theta}\)

Note by Sharky Kesa
1 year, 10 months ago

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Here's a nice result which involves all six ratios :

\[(\sin \theta + \cos \theta)(\tan \theta + \cot \theta)=\sec \theta + \csc \theta\] Nihar Mahajan · 1 year, 10 months ago

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Can you hear the Proving Trigonometric Identities Wiki page calling out your name? Calvin Lin Staff · 1 year, 10 months ago

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Thank You, it is a good practice for we beginners! Swapnil Das · 1 year, 10 months ago

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