Trigonometry practice for beginners

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

sin⁡2θ+cos⁡2θ=1\sin^2{\theta} + \cos^2{\theta} = 1

unless otherwise stated. More will be written.

Prove the following:

  1. csc⁡θ⋅cos⁡θ=cot⁡θ\csc{\theta} \cdot \cos{\theta} = \cot{\theta}

  2. csc⁡θ⋅tan⁡θ=sec⁡θ\csc{\theta}\cdot\tan{\theta}=\sec{\theta}

  3. 1+tan⁡2(−θ)=sec⁡2θ1+\tan^2(-\theta) = \sec^2{\theta}

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

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

  6. tan⁡θ⋅cot⁡θ−cos⁡2θ=sin⁡2θ\tan{\theta}\cdot\cot{\theta}-\cos^2{\theta}=\sin^2{\theta}

  7. sin⁡θ⋅csc⁡θ−cos⁡2θ=sin⁡2θ\sin{\theta}\cdot\csc{\theta}-\cos^2{\theta}=\sin^2{\theta}

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

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

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

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

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

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

  14. tan⁡2θ⋅cos⁡2θ+cot⁡2θ⋅sin⁡2θ=1\tan^2{\theta}\cdot\cos^2{\theta}+\cot^2{\theta}\cdot\sin^2{\theta}=1

  15. sec⁡4θ−sec⁡2θ=tan⁡4θ+tan⁡2θ\sec^4{\theta}-\sec^2{\theta}=\tan^4{\theta}+\tan^2{\theta}

  16. sec⁡θ−tan⁡θ=cos⁡θ1+sin⁡θ\sec{\theta}-\tan{\theta}=\dfrac{\cos{\theta}}{1+\sin{\theta}}

  17. csc⁡θ−cot⁡θ=sin⁡θ1+sin⁡θ\csc{\theta}-\cot{\theta}=\dfrac{\sin{\theta}}{1+\sin{\theta}}

  18. 3sin⁡2θ+4cos⁡2θ=3+cos⁡2θ3\sin^2{\theta}+4\cos^2{\theta}=3+\cos^2{\theta}

  19. 9sec⁡2θ−5tan⁡2θ=5+4sec⁡2θ9\sec^2{\theta}-5\tan^2{\theta}=5+4\sec^2{\theta}

  20. 1−cos⁡2θ1+sin⁡θ=sin⁡θ1-\dfrac{\cos^2{\theta}}{1+\sin{\theta}}=\sin{\theta}

  21. 1−sin⁡2θ1−cos⁡θ=−cos⁡θ1-\dfrac{\sin^2{\theta}}{1-\cos{\theta}}=-\cos{\theta}

  22. 1+tan⁡θ1−tan⁡θ=cot⁡θ+1cot⁡θ−1\dfrac{1+\tan{\theta}}{1-\tan{\theta}}=\dfrac{\cot{\theta}+1}{\cot{\theta}-1}

  23. sec⁡θcsc⁡θ+sin⁡θcos⁡θ=2tan⁡θ\dfrac{\sec{\theta}}{\csc{\theta}}+\dfrac{\sin{\theta}}{\cos{\theta}}=2\tan{\theta}

  24. csc⁡θ−1cot⁡θ=cot⁡θcsc⁡θ+1\dfrac{\csc{\theta}-1}{\cot{\theta}}=\dfrac{\cot{\theta}}{\csc{\theta}+1}

  25. 1+sin⁡θ1−sin⁡θ=csc⁡θ+1csc⁡θ−1\dfrac{1+\sin{\theta}}{1-\sin{\theta}}=\dfrac{\csc{\theta}+1}{\csc{\theta}-1}

  26. 1−sin⁡θcos⁡θ+cos⁡θ1−sin⁡θ=2sec⁡θ\dfrac{1-\sin{\theta}}{\cos{\theta}}+\dfrac{\cos{\theta}}{1-\sin{\theta}}=2\sec{\theta}

  27. sin⁡θsin⁡θ−cos⁡θ=11−cot⁡θ\dfrac{\sin{\theta}}{\sin{\theta}-\cos{\theta}}=\dfrac{1}{1-\cot{\theta}}

  28. 1−sin⁡2θ1+cos⁡θ=cos⁡θ1-\dfrac{\sin^2{\theta}}{1+\cos{\theta}}=\cos{\theta}

  29. 1−sin⁡θ1+sin⁡θ=(sec⁡θ−tan⁡θ)2\dfrac{1-\sin{\theta}}{1+\sin{\theta}}=(\sec{\theta}-\tan{\theta})^2

  30. cos⁡θ1−tan⁡θ+sin⁡θ1−cot⁡θ=sin⁡θ+cos⁡θ\dfrac{\cos{\theta}}{1-\tan{\theta}}+\dfrac{\sin{\theta}}{1-\cot{\theta}}=\sin{\theta}+\cos{\theta}

  31. cot⁡θ1−tan⁡θ+tan⁡θ1−cot⁡θ=1+tan⁡θ+cot⁡θ\dfrac{\cot{\theta}}{1-\tan{\theta}}+\dfrac{\tan{\theta}}{1-\cot{\theta}}=1+\tan{\theta}+\cot{\theta}

  32. tan⁡θ+cos⁡θ1+sin⁡θ=sec⁡θ\tan{\theta}+\dfrac{\cos{\theta}}{1+\sin{\theta}}=\sec{\theta}

#Geometry #TrigonometricIdentities #Sharky

Note by Sharky Kesa
6 years, 1 month ago

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Comments

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

(sin⁡θ+cos⁡θ)(tan⁡θ+cot⁡θ)=sec⁡θ+csc⁡θ(\sin \theta + \cos \theta)(\tan \theta + \cot \theta)=\sec \theta + \csc \theta

Nihar Mahajan - 6 years, 1 month ago

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

Calvin Lin Staff - 6 years, 1 month ago

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

Swapnil Das - 6 years, 1 month ago

Log in to reply

yes

Ram Mohith - 3 years, 3 months ago
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