Trigonometry practice for beginners

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

sin2θ+cos2θ=1\sin^2{\theta} + \cos^2{\theta} = 1

unless otherwise stated. More will be written.

Prove the following:

  1. cscθcosθ=cotθ\csc{\theta} \cdot \cos{\theta} = \cot{\theta}

  2. cscθtanθ=secθ\csc{\theta}\cdot\tan{\theta}=\sec{\theta}

  3. 1+tan2(θ)=sec2θ1+\tan^2(-\theta) = \sec^2{\theta}

  4. cosθ(tanθ+cotθ)=cscθ\cos{\theta}(\tan{\theta}+\cot{\theta})=\csc{\theta}

  5. sinθ(tanθ+cotθ)=secθ\sin{\theta}(\tan{\theta}+\cot{\theta})=\sec{\theta}

  6. tanθcotθcos2θ=sin2θ\tan{\theta}\cdot\cot{\theta}-\cos^2{\theta}=\sin^2{\theta}

  7. sinθcscθcos2θ=sin2θ\sin{\theta}\cdot\csc{\theta}-\cos^2{\theta}=\sin^2{\theta}

  8. (secθ1)(secθ+1)=tan2θ(\sec{\theta}-1)(\sec{\theta}+1)=\tan^2{\theta}

  9. (cscθ1)(cscθ+1)=cot2θ(\csc{\theta}-1)(\csc{\theta}+1)=\cot^2{\theta}

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

  11. sin2θ(1+cot2θ)=1\sin^2{\theta}(1+\cot^2{\theta})=1

  12. (1sin2θ)(1+tanθ)=1(1-\sin^2{\theta})(1+\tan{\theta})=1

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

  14. tan2θcos2θ+cot2θsin2θ=1\tan^2{\theta}\cdot\cos^2{\theta}+\cot^2{\theta}\cdot\sin^2{\theta}=1

  15. sec4θsec2θ=tan4θ+tan2θ\sec^4{\theta}-\sec^2{\theta}=\tan^4{\theta}+\tan^2{\theta}

  16. secθtanθ=cosθ1+sinθ\sec{\theta}-\tan{\theta}=\dfrac{\cos{\theta}}{1+\sin{\theta}}

  17. cscθcotθ=sinθ1+sinθ\csc{\theta}-\cot{\theta}=\dfrac{\sin{\theta}}{1+\sin{\theta}}

  18. 3sin2θ+4cos2θ=3+cos2θ3\sin^2{\theta}+4\cos^2{\theta}=3+\cos^2{\theta}

  19. 9sec2θ5tan2θ=5+4sec2θ9\sec^2{\theta}-5\tan^2{\theta}=5+4\sec^2{\theta}

  20. 1cos2θ1+sinθ=sinθ1-\dfrac{\cos^2{\theta}}{1+\sin{\theta}}=\sin{\theta}

  21. 1sin2θ1cosθ=cosθ1-\dfrac{\sin^2{\theta}}{1-\cos{\theta}}=-\cos{\theta}

  22. 1+tanθ1tanθ=cotθ+1cotθ1\dfrac{1+\tan{\theta}}{1-\tan{\theta}}=\dfrac{\cot{\theta}+1}{\cot{\theta}-1}

  23. secθcscθ+sinθcosθ=2tanθ\dfrac{\sec{\theta}}{\csc{\theta}}+\dfrac{\sin{\theta}}{\cos{\theta}}=2\tan{\theta}

  24. cscθ1cotθ=cotθcscθ+1\dfrac{\csc{\theta}-1}{\cot{\theta}}=\dfrac{\cot{\theta}}{\csc{\theta}+1}

  25. 1+sinθ1sinθ=cscθ+1cscθ1\dfrac{1+\sin{\theta}}{1-\sin{\theta}}=\dfrac{\csc{\theta}+1}{\csc{\theta}-1}

  26. 1sinθcosθ+cosθ1sinθ=2secθ\dfrac{1-\sin{\theta}}{\cos{\theta}}+\dfrac{\cos{\theta}}{1-\sin{\theta}}=2\sec{\theta}

  27. sinθsinθcosθ=11cotθ\dfrac{\sin{\theta}}{\sin{\theta}-\cos{\theta}}=\dfrac{1}{1-\cot{\theta}}

  28. 1sin2θ1+cosθ=cosθ1-\dfrac{\sin^2{\theta}}{1+\cos{\theta}}=\cos{\theta}

  29. 1sinθ1+sinθ=(secθtanθ)2\dfrac{1-\sin{\theta}}{1+\sin{\theta}}=(\sec{\theta}-\tan{\theta})^2

  30. cosθ1tanθ+sinθ1cotθ=sinθ+cosθ\dfrac{\cos{\theta}}{1-\tan{\theta}}+\dfrac{\sin{\theta}}{1-\cot{\theta}}=\sin{\theta}+\cos{\theta}

  31. cotθ1tanθ+tanθ1cotθ=1+tanθ+cotθ\dfrac{\cot{\theta}}{1-\tan{\theta}}+\dfrac{\tan{\theta}}{1-\cot{\theta}}=1+\tan{\theta}+\cot{\theta}

  32. tanθ+cosθ1+sinθ=secθ\tan{\theta}+\dfrac{\cos{\theta}}{1+\sin{\theta}}=\sec{\theta}

Note by Sharky Kesa
4 years, 3 months ago

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1 vote

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

(sinθ+cosθ)(tanθ+cotθ)=secθ+cscθ(\sin \theta + \cos \theta)(\tan \theta + \cot \theta)=\sec \theta + \csc \theta

Nihar Mahajan - 4 years, 3 months ago

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Can you hear the Proving Trigonometric Identities Wiki page calling out your name?

Calvin Lin Staff - 4 years, 3 months ago

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Thank You, it is a good practice for we beginners!

Swapnil Das - 4 years, 3 months ago

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yes

Ram Mohith - 1 year, 5 months ago

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