Question

1-tan^2θ/ cot^2θ -1 = tan^2θ

Answer 1

Answer :

L.H.S. = ( 1 - tan²θ ) / ( cot²θ - 1 )

= ( 1 - tan²θ ) / { ( 1 / tan²θ ) - 1 }

= ( 1 - tan²θ ) / { ( 1 - tan²θ ) / tan²θ }

= tan²θ

= R.H.S.

Hence, proved.

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Answer 2

SOLUTION:-

We have,

$\frac{1 - {tan}^{2} \theta }{ {cot}^{2} \theta - 1 } = tan {}^{2} \theta$

Take L.H.S

$= > ( \frac{1 - {tan}\theta }{1 - {cot} \theta} ) {}^{2} \\ \\ = > ( \frac{1 - \frac{sin \theta}{cos \theta} }{1 - \frac{cos \theta}{sin \theta} } ) {}^{2} \\ \\ = > ( \frac{ \frac{cos \theta - sin \theta}{cos \theta} }{ \frac{sin \theta - cos \theta}{sin \theta} } ) {}^{2} \\ \\ = > ( \frac{ \frac{ - (sin \theta - cos \theta)}{cos \theta} }{ \frac{sin \theta - cos \theta}{sin \theta} } ) {}^{2} \\ \\ = > ( \frac{ - (sin \theta - cos \theta)}{sin \theta - cos \theta} \times \frac{sin \theta}{cos \theta} ) {}^{2} \\ \\ = > ( - \frac{sin \theta}{cos \theta} ) {}^{2} \\ \\ = > \frac{ {sin}^{2} \theta }{ {cos}^{2} \theta } \\ \\ = > {tan}^{2} \theta \: \: \: \: \: \: \: \: \: [R.H.S]$

[Proved].

Thank you