prove trigonometry question
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tarun0001:
divide LHS by cosx in numerator and denominator then rationalise u get your ans
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Answered by
2
Divide lhs by cosx
Secx + tanx) /( secx - tanx)
( secx +tanx) ^2/ sec^2x - tan^2x
( secx +tanx) ^2 rhs ....... Proved
Secx + tanx) /( secx - tanx)
( secx +tanx) ^2/ sec^2x - tan^2x
( secx +tanx) ^2 rhs ....... Proved
Answered by
5
Theta is written as A ,
LHS = >


RHS = >


LHS = RHS
Hence, proved.
LHS = >
RHS = >
LHS = RHS
Hence, proved.
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