Question

prove that


[1+tan²x/1+cot²x]- [1-tan²x/1-cot²x]²-0

pls answer fast​

Answer 1

↣Correct Question:-

Prove that

$\sf\Bigg(\dfrac{1 + {tan}^{2} x}{1 + {cot}^{2}x } \Bigg) -\Bigg(\dfrac{1 - tan \: x}{1 - {cot }\: x }\Bigg) ^{2} = 0$

↣Identities used

$\blue{\sf \longrightarrow \: 1 + {tan}^{2} x = sec^{2} x \: \: \: - (1)}$

$\red{\sf \longrightarrow \: 1 + {cot}^{2} x = cosec^{2} x \: \: - (2)}$

$\orange{ \sf \longrightarrow \:tan \: x = \dfrac{sin \: x}{cos \: x} \: \: \: - (3)}$

↣Solution:-

We have LHS

$\implies\sf\Bigg(\dfrac{1 + {tan}^{2} x}{1 + {cot}^{2}x } \Bigg) -\Bigg(\dfrac{1 - tan \: x}{1 - {cot} \: x }\Bigg) ^{2}$

$\sf \implies \Bigg( \dfrac{sec^{2} \: x }{ {cosec}^{2} \: x} \Bigg) - \dfrac{\Bigg(1 - \frac{sin \: x}{cos \: x} \Bigg)^{2} }{\Bigg(1 - \frac{cos \: x}{sin \: x } \Bigg)^{2} }$

$\sf \implies \Bigg(\dfrac{1}{ {cos}^{2}x } \times sin ^{2} x\Bigg) - \Bigg[\dfrac{ \dfrac{(cos \: x - sin \: x)}{ {cos}^{2} x} }{ \dfrac{(sin \: x - cos \: x)}{ {sin}^{2}x } } \Bigg]$

$\sf \implies\Bigg( \dfrac{ {sin}^{2}x }{ {cos}^{2} x} \Bigg) - \Bigg(\dfrac{ \cancel{cos \: x - sin \: x}}{ {cos}^{2}x } \times \dfrac{ {sin}^{2} x} {\cancel{cos \: x - sin \: x} }\Bigg)$

$\sf \implies\Bigg( \dfrac{ {sin}^{2}x }{ {cos}^{2} x} \Bigg) - \Bigg( \dfrac{ {sin}^{2}x }{ {cos}^{2} x} \Bigg)$

$\implies \sf \: {tan}^{2} x - {tan}^{2} x$

$\sf \implies 0$

$\blue{ \implies \sf \mathbb{R} . \mathbb{H}.\mathbb{S}}$

Hence, proved