Question

to prove ((cot A - 1)/(2-sec^2 A))=((cot A )÷(1+tan A))

Answer 1

Hi...☺

Here is your answer...✌

LHS

$= \frac{ \cot( \alpha ) - 1 }{2 - \sec {}^{2} ( \alpha ) } \\ \\ = \frac{ \cot( \alpha ) - \tan( \alpha ) \cot( \alpha ) }{2 - (1 + \tan {}^{2} ( \alpha )) } \: \: \\ \\ = \frac{\cot( \alpha ) (1 - \tan( \alpha ) )}{2 - 1 - \tan {}^{2} ( \alpha ) } \\ \\ = \frac{ \cot( \alpha)(1 - \tan( \alpha ) ) }{1 - \tan {}^{2} ( \alpha ) } \\ \\ = \frac{ \cot( \alpha) (1 - \tan( \alpha ) ) }{(1 + \tan( \alpha ))( 1 - \tan( \alpha )) } \\ \\ = \frac{ \cot( \alpha ) }{1 + \tan( \alpha ) }$

= RHS [ Hence Proved ]

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★Formula used in 2nd step :

=> tanA × cotA = 1

=> sec²A = 1 + tan²A

Hope it helps ✌