MCQ
$\frac{\cos10^\circ+\sin10^\circ}{\cos10^\circ-\sin10^\circ}$ is equal to:
  • $\tan55^\circ$
  • B
    $\cot55^\circ$
  • C
    $-\tan35^\circ$
  • D
    $-\cot35^\circ$

Answer

Correct option: A.
$\tan55^\circ$
$\frac{\cos10^\circ+\sin10^\circ}{\cos10^\circ-\sin10^\circ}$
$=\frac{1+\tan10^\circ}{1-\tan10^\circ}$ $[$Dividing the numerator and denominator by $\cos10^\circ]$
$=\frac{\tan45^\circ+\tan10^\circ}{1-\tan45^\circ\times\tan10^\circ}$
$=\tan(45^\circ+10^\circ)$ $\Big[$Using $\tan(\text{A+B})=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}\Big]$
$=\tan55^\circ$

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