cosx=root99\10
find the value of log sinx+logcosx+logtanx
rishimhaske:
is it root 99 / 10 or root 99/root 10
root99\10
can u write it in words
Answers
Answered by
7
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
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
sorry sir but logsinxcosxtanx=logsin^2x
dear sir can u plss check both once
Answered by
4
log sinx+logcosx+logtanx
=logsinx*cosx*tanx
=logsinx*cosx*(six/cosx)
=logsin^2x
=log(1-cos^2x)
=log(1-99/100)
=log(1/100)
=logsinx*cosx*tanx
=logsinx*cosx*(six/cosx)
=logsin^2x
=log(1-cos^2x)
=log(1-99/100)
=log(1/100)
oh the root is only for 99 right sir
is it right sir
dont say me sir.... iam student.....bro
yes
ohhhhhhhhhhhhh
is it right
yes
wr is log in ur answr then
i have edited it,.....sory
kk
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