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# Answer to Question #3669 in Trigonometry for Abhijith K U

Question #3669
Prove the following: 1) (1-sin^2 Q) sec^2 Q=1 2) cos^2Q (1+tan^2Q)=1 3) cos^2Q + 1/1+cot^2 Q=1 4) 1/1+sinQ + 1/1-sinQ=2sec^2Q
<img src="/cgi-bin/mimetex.cgi?1%29%20%281-%5Csin%5E2%20Q%29%20%5Csec%5E2%20Q=1%20%5C%5C%20%5Ccos%5E2%20Q%20*%201/%5Ccos%5E2%20Q%20=%201%20%5C%5C%202%29%20%5Ccos%5E2Q%20%281+%5Ctan%5E2Q%29=1%20%5C%5C%20%5Ccos%5E2%20Q%20%281%20+%20%5Csin%5E2%20Q%20/%20%5Ccos%5E2%20Q%29%20=%20%5Ccos%5E2%20Q%20%5Cfrac%7B%5Ccos%5E2%20Q%20+%20%5Csin%5E2%20Q%7D%7B%5Ccos%5E2%20Q%7D%20=%20%5C%5C%20=%20%5Ccos%5E2%20Q%20+%20%5Csin%5E2%20Q%20=%201%20%5C%5C%203%29%20%5Ccos%5E2Q%20+%20%5Cfrac%7B1%7D%7B1+%5Ccot%5E2%20Q%7D=1%20%5C%5C%20%5Ccos%5E2Q%20+%20%5Cfrac%7B1%7D%7B1+%5Ccos%5E2%20Q%20/%5Csin%5E2%20Q%20%7D%20=%20%5Ccos%5E2Q%20+%20%5Cfrac%7B%5Csin%5E2%20Q%7D%7B%5Csin%5E2%20Q+%5Ccos%5E2%20Q%20%7D%20%5C%5C%20=%20%5Ccos%5E2Q%20+%20%5Csin%5E2%20Q%20=%201%5C%5C%204%29%20%5Cfrac%7B1%7D%7B1+%5Csin%20Q%7D%20+%20%5Cfrac%7B1%7D%7B1-%5Csin%20Q%7D=2%5Csec%5E2Q%20%5C%5C%20%5Cfrac%7B1+%5Csin%20Q%20+%201%20-%20%5Csin%20Q%7D%7B%281+%5Csin%20Q%29%281-%5Csin%20Q%29%7D%20=%20%5Cfrac%7B2%7D%7B1%20-%20%5Csin%5E2%20Q%7D%20=%20%5C%5C%20=%20%5Cfrac%7B2%7D%7B%20%5Ccos%5E2%20Q%7D%20=%202%5Csec%5E2Q" title="1) (1-\sin^2 Q) \sec^2 Q=1 \\ \cos^2 Q * 1/\cos^2 Q = 1 \\ 2) \cos^2Q (1+\tan^2Q)=1 \\ \cos^2 Q (1 + \sin^2 Q / \cos^2 Q) = \cos^2 Q \frac{\cos^2 Q + \sin^2 Q}{\cos^2 Q} = \\ = \cos^2 Q + \sin^2 Q = 1 \\ 3) \cos^2Q + \frac{1}{1+\cot^2 Q}=1 \\ \cos^2Q + \frac{1}{1+\cos^2 Q /\sin^2 Q } = \cos^2Q + \frac{\sin^2 Q}{\sin^2 Q+\cos^2 Q } \\ = \cos^2Q + \sin^2 Q = 1\\ 4) \frac{1}{1+\sin Q} + \frac{1}{1-\sin Q}=2\sec^2Q \\ \frac{1+\sin Q + 1 - \sin Q}{(1+\sin Q)(1-\sin Q)} = \frac{2}{1 - \sin^2 Q} = \\ = \frac{2}{ \cos^2 Q} = 2\sec^2Q">

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