Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions
Let 0 < x < π...
Question
Let 0<x<π4, then the value of (sec2x−tan2x) is
A
tan(x−π4)
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B
tan(π4−x)
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C
tan(x+π4)
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D
tan2(x+π4)
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Solution
The correct option is Btan(π4−x) sec2x−tan2x=1−sin2xcos2x=1−2sinxcosxcos2x−sin2x=(cosx−sinx)2(cosx−sinx)(cosx+sinx)=cosx−sinxcosx+sinx(∵cosx≠sinx⇒cosx−sinx≠0)=1−tanx1+tanx(∵cosx≠0) =tan(π4−x)