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\(\frac{1-tan^2\,45°}{1+tan^2\,45°}\) = (a) tan 90° (b) 1 (c) sin 45° (d) 0 |
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Answer» (d) 0 \(\frac{1-tan^2\,45°}{1+tan^2\,45°}\) = \(\frac{1-1}{1+1}= \frac02=0\) (∵ tan 45° = 1) |
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