Trig2.15

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Prove that\tan (\theta-\frac{3\pi}{4}) \tan (\frac{7\pi}{4}+\theta)+1=0\,

LHS=\tan (\theta-135) \tan (315+\theta)+1\,

\frac{\tan \theta-\tan 135}{1+\tan \theta \tan 135} \cdot \frac{\tan 315+\tan \theta}{1-\tan 315 \tan \theta}+1\,

\frac{\tan \theta+1}{1-\tan \theta}\cdot \frac{-1+\tan \theta}{1+\tan \theta}+1\,

-1+1=0\,=RHS.


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