Answers

2014-11-04T16:50:23+05:30
We know
logA+logB+logC=logABC
logSin+logCos+logTan=log  sin cos tan
but
tan=sin/cos
so
log sin cos tan=logsin cos (sin cos)
log sin cos tan=sin²
by identity,
sin²=1-cos²
so
log sin cos tan=log1-(√99/10)²
log sin cos tan=log1/100
so answer is
log1/100

1 5 1
sorry sir but logsinxcosxtanx=logsin^2x
dear sir can u plss check both once
2014-11-04T16:53:05+05:30
Log sinx+logcosx+logtanx
=logsinx*cosx*tanx
=logsinx*cosx*(six/cosx)
=logsin^2x
=log(1-cos^2x)
=log(1-99/100)
=log(1/100)
0
is it right
yes
wr is log in ur answr then
i have edited it,.....sory
kk