prove that
[1+tan²x/1+cot²x]- [1-tan²x/1-cot²x]²-0
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prove that
[1+tan²x/1+cot²x]- [1-tan²x/1-cot²x]²-0
pls answer fast
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↣Correct Question:-
Prove that
[tex] \sf\Bigg(\dfrac{1 + {tan}^{2} x}{1 + {cot}^{2}x } \Bigg) -\Bigg(\dfrac{1 - tan \: x}{1 - {cot }\: x }\Bigg) ^{2} = 0[/tex]
↣Identities used
[tex] \blue{\sf \longrightarrow \: 1 + {tan}^{2} x = sec^{2} x \: \: \: - (1)}[/tex]
[tex] \red{\sf \longrightarrow \: 1 + {cot}^{2} x = cosec^{2} x \: \: - (2)}[/tex]
[tex] \orange{ \sf \longrightarrow \:tan \: x = \dfrac{sin \: x}{cos \: x} \: \: \: - (3)}[/tex]
↣Solution:-
We have LHS
[tex] \implies\sf\Bigg(\dfrac{1 + {tan}^{2} x}{1 + {cot}^{2}x } \Bigg) -\Bigg(\dfrac{1 - tan \: x}{1 - {cot} \: x }\Bigg) ^{2} [/tex]
[tex] \sf \implies \Bigg( \dfrac{sec^{2} \: x }{ {cosec}^{2} \: x} \Bigg) - \dfrac{\Bigg(1 - \frac{sin \: x}{cos \: x} \Bigg)^{2} }{\Bigg(1 - \frac{cos \: x}{sin \: x } \Bigg)^{2} } [/tex]
[tex] \sf \implies \Bigg(\dfrac{1}{ {cos}^{2}x } \times sin ^{2} x\Bigg) - \Bigg[\dfrac{ \dfrac{(cos \: x - sin \: x)}{ {cos}^{2} x} }{ \dfrac{(sin \: x - cos \: x)}{ {sin}^{2}x } } \Bigg][/tex]
[tex] \sf \implies\Bigg( \dfrac{ {sin}^{2}x }{ {cos}^{2} x} \Bigg) - \Bigg(\dfrac{ \cancel{cos \: x - sin \: x}}{ {cos}^{2}x } \times \dfrac{ {sin}^{2} x} {\cancel{cos \: x - sin \: x} }\Bigg)[/tex]
[tex]\sf \implies\Bigg( \dfrac{ {sin}^{2}x }{ {cos}^{2} x} \Bigg) - \Bigg( \dfrac{ {sin}^{2}x }{ {cos}^{2} x} \Bigg) [/tex]
[tex] \implies \sf \: {tan}^{2} x - {tan}^{2} x[/tex]
[tex] \sf \implies 0[/tex]
[tex] \blue{ \implies \sf \mathbb{R} . \mathbb{H}.\mathbb{S}}[/tex]
Hence, proved