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[tex]to \: find : \\ \frac{1 + tan {}^{2}x }{1 + cot {}^{2} x} \: - [ \frac{1 - tan \: x}{1 - cot \: x} ] {}^{2} \\ \\ = \frac{1 + tan {}^{2}x }{1 + cot {}^{2} x} - [ \frac{1 + tan {}^{2}x - 2.tan \: x }{1 + cot {}^{2} x - 2.cot \: x} ] \\ \\ let \: assume \: that \\ 1 + tan {}^{2} x = a \\ 1 + cot {}^{2} x = b \\ \\ \frac{a}{b} - [ \frac{a - 2.tan \: x}{b - 2.cot \: x} ] \\ \\ = \frac{a(b - 2.cot \: x) - b(a - 2.tan \: x)}{b(b - 2.cot \: x)} \\ \\ = \frac{ab - 2a.cot \: x - ab + 2b.tan \: x}{b {}^{2} - 2b.cot \: x} \\ \\ = \frac{2(b.tan \: x - a.cot \: x)}{b {}^{2} - 2b.tan \: x} \\ \\ let \: the \: denominator \: be \: k \\ thus \: then \: accordingly \\ and \: resubstituting \: values \\ we \: get \\ \\ = \frac{2[(1 + cot{}^{2}x)(tan \: x) - (1 + tan {}^{2} x)(cot\: x) ]}{k} \\ \\ = \frac{2}{k} (tan \: x + cot {}^{2} x.tan \: x - cot \: x - tan {}^{2} x.cot \: x) \\ \\ = \frac{2}{k} (tan \: x + cot \: x - cot \: x - tan \: x) \\ \\ = \frac{2}{k} \times 0 \\ \\ = 0[/tex]
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Step-by-step explanation:
[tex]to \: find : \\ \frac{1 + tan {}^{2}x }{1 + cot {}^{2} x} \: - [ \frac{1 - tan \: x}{1 - cot \: x} ] {}^{2} \\ \\ = \frac{1 + tan {}^{2}x }{1 + cot {}^{2} x} - [ \frac{1 + tan {}^{2}x - 2.tan \: x }{1 + cot {}^{2} x - 2.cot \: x} ] \\ \\ let \: assume \: that \\ 1 + tan {}^{2} x = a \\ 1 + cot {}^{2} x = b \\ \\ \frac{a}{b} - [ \frac{a - 2.tan \: x}{b - 2.cot \: x} ] \\ \\ = \frac{a(b - 2.cot \: x) - b(a - 2.tan \: x)}{b(b - 2.cot \: x)} \\ \\ = \frac{ab - 2a.cot \: x - ab + 2b.tan \: x}{b {}^{2} - 2b.cot \: x} \\ \\ = \frac{2(b.tan \: x - a.cot \: x)}{b {}^{2} - 2b.tan \: x} \\ \\ let \: the \: denominator \: be \: k \\ thus \: then \: accordingly \\ and \: resubstituting \: values \\ we \: get \\ \\ = \frac{2[(1 + cot{}^{2}x)(tan \: x) - (1 + tan {}^{2} x)(cot\: x) ]}{k} \\ \\ = \frac{2}{k} (tan \: x + cot {}^{2} x.tan \: x - cot \: x - tan {}^{2} x.cot \: x) \\ \\ = \frac{2}{k} (tan \: x + cot \: x - cot \: x - tan \: x) \\ \\ = \frac{2}{k} \times 0 \\ \\ = 0[/tex]