First slide

Trigonometric Identities

Question

If $b > 1, \sin t > 0, \cos t > 0 and \log_{b} (\sin t) = x, then \log_{b} (\cos t) =$

Moderate

A

$\frac{1}{2} \log_{b} (1 - b^{2 x})$
B

$2 \log (1 - b^{x / 2})$
C

$\log_{b} \sqrt{1 - b^{2 x}}$
D

$\sqrt{1 - x^{2}}$

Solution

$\begin{array}{l} \log_{b} \sin t = x or \sin t = b^{x} \\ Let \log_{b} (\cos t) = y, then b^{y} = \cos t \\ or b^{2 y} = \cos^{2} t = 1 - \sin^{2} t = 1 - b^{2 x} \\ or 2 y = \log_{b} (1 - b^{2 x}) \\ or y = \frac{1}{2} \log_{b} (1 - b^{2 x}) = \log_{b} \sqrt{1 - b^{2 x}} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App