(cos 10° – sin 10°)/(cos 10° + sin 10°) is equal to (a) tan 55° (b) cot 55° (c) – tan 35° (d)

Find the Value of (cos 10° − sin 10°)/(cos 10° + sin 10°)

Question:
\[ \frac{\cos10^\circ-\sin10^\circ} {\cos10^\circ+\sin10^\circ} \] is equal to

(a) \(\tan55^\circ\)
(b) \(\cot55^\circ\)
(c) \(-\tan35^\circ\)
(d) \(-\cot35^\circ\)

Solution

Divide numerator and denominator by

\[ \cos10^\circ \]

Then,

\[ \frac{\cos10^\circ-\sin10^\circ} {\cos10^\circ+\sin10^\circ} = \frac{1-\tan10^\circ} {1+\tan10^\circ} \]

Using the identity:

\[ \tan(A-B) = \frac{\tan A-\tan B} {1+\tan A\tan B} \]

Take

\[ A=45^\circ, \qquad B=10^\circ \]

Since

\[ \tan45^\circ=1 \]

we get

\[ \tan(45^\circ-10^\circ) = \frac{1-\tan10^\circ} {1+\tan10^\circ} \]

\[ = \tan35^\circ \]

Therefore,

\[ \frac{\cos10^\circ-\sin10^\circ} {\cos10^\circ+\sin10^\circ} = \tan35^\circ \]

Now,

\[ \cot55^\circ = \tan(90^\circ-55^\circ) = \tan35^\circ \]

Hence,

\[ \boxed{ \frac{\cos10^\circ-\sin10^\circ} {\cos10^\circ+\sin10^\circ} = \cot55^\circ } \]

\[ \boxed{ \frac{\cos10^\circ-\sin10^\circ} {\cos10^\circ+\sin10^\circ} = \cot55^\circ } \]

Correct Option: (b)

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