Question

if cosx=√99/10,then find the value of log(sinx)+ log(tanx) (0<x>90)

Answer 1

log(sinx)+log(tanx)=log(sin^2x)-logcosx
=log(1/10)-log(√99/100)

Answer 2

Answer:

The required value of the given expression is -2

Step-by-step explanation:

$\cos x=\frac{\sqrt{99}}{10}\\\\Now,\log(\sin x)+\log(\tan x)\\\\=\log(\sin x)+\log(\frac{\sin x}{\cos x})\\\\=\log(\sin x)+\log(\sin x)-\log(\cos x)\\\\=2\log(\sin x)-\log(\cos x)\\\\=\log(\sin^2x)-\log(\cos x)\\\\=\log(1-\cos^2x)-\log(\cos x)\\\\=\log(1-\frac{99}{100})-\log(\frac{\sqrt{99}}{10})\\\\=\log(\frac{1}{100})-\log(\frac{\sqrt{99}}{10})\\\\=\log 1-\log 100-\log 0.99\\\\\approx -2$

The required value of the given expression is -2