If tan 69° + tan 66° − tan 69° tan 66° = 2k, Find k

Solution

Using the identity:

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

Take

\[ A=69^\circ, \qquad B=66^\circ \]

Then,

\[ A+B=135^\circ \]

Therefore,

\[ \tan135^\circ = \frac{ \tan69^\circ+\tan66^\circ } { 1-\tan69^\circ\tan66^\circ } \]

Since

\[ \tan135^\circ=-1 \]

we get

\[ -1 = \frac{ \tan69^\circ+\tan66^\circ } { 1-\tan69^\circ\tan66^\circ } \]

Cross multiplying,

\[ -1+\tan69^\circ\tan66^\circ = \tan69^\circ+\tan66^\circ \]

Rearranging,

\[ \tan69^\circ+\tan66^\circ-\tan69^\circ\tan66^\circ = -1 \]

Given,

\[ \tan69^\circ+\tan66^\circ-\tan69^\circ\tan66^\circ=2k \]

Hence,

\[ 2k=-1 \]

Therefore,

\[ k=-\frac{1}{2} \]

Final Answer

\[ \boxed{ k=-\frac{1}{2} } \]

Correct Option: (c)