Question

If sin x = $\frac{\sqrt{99}}{10}$ , then find the value of log (sin x) + log (tan x) (0 < x < 90).

Answer 1

$\begin{array}{c}Given\\ \sin x=\frac{\sqrt{99}}{10}\\ Now,\\ \log \left(\sin x\right)-\log \left(\tan x\right)\\ =\log \left(\sin x.\tan x\right)\therefore \left[\log a+\log b=\log \left(ab\right)\right]\\ =\log \left(\sin x.\frac{\sin x}{\cos x}\right)\\ =\log \left(\frac{{\sin }^{2}x}{\cos x}\right)----\left(1\right)\\ \sin x=\frac{\sqrt{99}}{10}\\ \therefore \cos x=\sqrt{1-{\sin }^{2}x}=\sqrt{1-\frac{99}{100}}\\ cosx=\sqrt{{\left(\frac{1}{10}\right)}^2}\\ \Rightarrow \cos x=\frac{1}{10}\left[0<x<{90}^{\circ }\right]\\ Substitutevalueof\sin xand\cos xineq{u}^n\left(1\right)weget\\ \log \left(\sin x\right)+\log \left(\tan x\right)\\ =\log \left(\frac{\frac{99}{100}}{\frac{1}{10}}\right)\\ =\log \left(\frac{99}{100}\times \frac{10}{1}\right)\\ =\log \left(\frac{99}{10}\right)\\ =\log 99-\log 10\\ =\log 99-1\end{array}$