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let theta = x
we have,
sinx+cosx = m
squaring on both side
sin²x+cos²+2sinxcosx = m²
1+2sinxcosx = m²
2sinxcosx = m²-1
and also we have, secx+cosecx = n
have to prove,,
n(m²-1) = 2m
LHS, n(m²-1)
(secx+cosecx)(2sinxcosx)
(1/cosx+1/sinx)(sin2x)
[(sinx+cosx)/cosxsinx](2sinxcosx)
2(sinx+cosx)
RHS, 2m
2(sinx+cosx)
So, RHS = LHS
Hence, proved
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