Question

if 0<x<90 and cosx=3/√10 then find the value of
log sinx+log cosx+log tanx

Answer 1

cos x = 3/√10

log(sinx) + log(cosx) + log(tanx)
= log(sinx × cosx × tanx )       [using log (a) +log (b) = log (ab)]
= log ( sinx × cosx ×(sinx/cosx))
=log (sinx × sinx)
=log (sin²x)

sin²x = 1 - cos²x = 1 - (3/√10)² = 1 - (9/10) = (10-9)/10 = 1/10

log(sin²x) = log(1/10) = -1