Question

If $b>1,\sin t>0,\cos t>0$ and ${\log }_b\left(\sin t\right)=x$, then ${\log }_b\left(\cos t\right)$ is equal to

• A. $\frac{1}{2}{\log }_b(1-b^{2x})$
• B. $2\log (1-b^{x/2})$
• C. ${\log }_b\sqrt{1-b^{2x}}{\log }_b(1-b^{2x})$
• D. $\sqrt{1-x^2}$

Answer 1

The correct option is A $\frac{1}{2}{\log }_b(1-b^{2x})$

Hence
$sint=b^x$
Or
$si{n}^{2}t=b^{2x}$
$1-co{s}^{2}t=b^{2x}$
$cost=\sqrt{1-b^{2x}}$
Or
$lo{g}_b\left(cost\right)=\frac{1}{2}lo{g}_b(1-b^{2x})$.