Question

$\frac{(\cos 9°+\sin 9°)}{(\cos 9°-\sin 9°)}$ is equal to

• A. $\tan 26°$
• B. $\tan 81°$
• C. $\tan 51°$
• D. $\tan 54°$
• E. $\tan 46°$

Answer 1

The correct option is D

$\tan 54°$

Explanation or the correct option:

Solve the expression $\frac{(\cos 9°+\sin 9°)}{(\cos 9°–\sin 9°)}$

By Divide the numerator and denominator by cos 9°, we get

$\frac{\left({\displaystyle \frac{\cos 9°}{\cos 9°}}+{\displaystyle \frac{\sin 9°}{\cos 9°}}\right)}{\left({\displaystyle \frac{\cos 9°}{\cos 9°}}–{\displaystyle \frac{\sin 9°}{\cos 9°}}\right)}$

$=\frac{(1+\tan 9°)}{(1–\tan 9°)}$

$=\frac{(\tan 45°+\tan 9°)}{[1–\tan 9°\times (1\left)\right]}$ ; $\left[\mathbf{\because }\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathbf{45}\mathbf{°}\mathbf{=}\mathbf{1}\right]$

$=\frac{(\tan 45°+\tan 9°)}{(1–\tan 9°\tan 45°)}$

$=\tan (45°+9°)$ ; $\left[\mathbf{\because }\frac{\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{A}\mathbf{+}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{B}}{\mathbf{1}\mathbf{-}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{A}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{B}}\mathbf{=}\mathit{t}\mathit{a}\mathit{n}\mathbf{(}\mathit{A}\mathbf{+}\mathit{B}\mathbf{)}\right]$

$=\mathbf{}\mathit{t}\mathit{a}\mathit{n}\mathbf{}\mathbf{54}\mathbf{°}$

Hence, Option ‘D’ is Correct.