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Question: Answered & Verified by Expert
$\frac{\cos 13^{\circ}-\sin 13^{\circ}}{\cos 13^{\circ}+\sin 13^{\circ}}+\frac{1}{\cot 148^{\circ}}$ is equal to
MathematicsTrigonometric EquationsTS EAMCETTS EAMCET 2016
Options:
  • A 1
  • B -1
  • C 0
  • D $\frac{1}{2}$
Solution:
2542 Upvotes Verified Answer
The correct answer is: 0
Given
$$
\begin{aligned}
& \frac{\cos 13^{\circ}-\sin 13^{\circ}}{\cos 13^{\circ}+\sin 13^{\circ}}+\frac{1}{\cot 148^{\circ}} \\
& =\frac{1-\tan 13^{\circ}}{1+\tan 13^{\circ}}+\tan 148^{\circ} \\
& =\frac{\tan 45^{\circ}-\tan 13^{\circ}}{1+\tan 45^{\circ} \tan 13^{\circ}}+\tan 148^{\circ}
\end{aligned}
$$
$$
\begin{aligned}
& =\tan \left(45^{\circ}-13^{\circ}\right)+\tan \left(180^{\circ}-32^{\circ}\right) \\
& =\tan 32^{\circ}-\tan 32^{\circ}=0
\end{aligned}
$$

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