fast
class 10
chapter 8or 9
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First ,
l=sec+tan
By taking RHS,
(l²+1)/2l
={sec+tan}²+1/{2(sec+tan)}
={sec+tan}²+(sec²-tan²)/{2(sec+tan)}
={sec+tan}²+(sec-tan)(sec+tan)/{2(sec+tan)}
By taking sec+tan common,we get
sec+tan[(sec+tan)+(sec-tan)]/2(sec+tan)
(sec+tan)+(sec-tan)/2 (by canceling sec+tan)
2sec/2
=sec=LHS
adityakumar5155:
Sorry for late , this was a little long
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