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Prove(sec theta-tan theta)(sec theta+tan theta)=1​

User AdrianHHH
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1 Answer

25 votes
25 votes

Answer:


( \sec \theta - \tan \theta)( \sec \theta + \tan \theta)

from difference of two squares:


{ \boxed{ \tt{( {a}^(2) - {b}^(2)) = (a - b)(a + b) }}}


= { \sec }^(2) \theta - { \tan}^(2) \theta \\ = (1 + { \tan}^(2) \theta) - { \tan}^(2) \theta \\ = 1

hence proved

User Jerica
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