Question

tan 69°t tan 66°
l-tan baitan 66°​

Answer 1

Answer:

We know that

\begin{gathered}\begin{aligned} \tan ( A + B ) & = \frac { \tan A + \tan B } { 1 - \tan A \tan B } \\\\ \tan \left( 69 ^ { \circ } + 66 ^ { \circ } \right) & = \frac { \tan 69 ^ { \circ } + \tan 66 ^ { \circ } } { 1 - \tan 69 ^ { \circ } \tan 66 ^ { \circ } } \\\\ \tan 135 ^ { \circ } & = \frac { \tan 69 ^ { \circ } + \tan 66 ^ { \circ } } { 1 - \tan 69 ^ { \circ } \tan 66 ^ { \circ } } \end{aligned}\end{gathered}tan(A+B)tan(69∘+66∘)tan135∘=1−tanAtanBtanA+tanB=1−tan69∘tan66∘tan69∘+tan66∘=1−tan69∘tan66∘tan69∘+tan66∘

We know \tan 135^{\circ} =-1tan135∘=−1 since \tan \frac{3\pi}{4} =-1tan43π=−1

- 1 = \frac { \tan 69 ^ { \circ } + \tan 66 ^ { \circ } } { 1 - \tan 69 ^ { \circ } \tan 66 ^ { \circ } }−1=1−tan69∘tan66∘tan69∘+tan66∘

\tan 69 ^ { \circ } + \tan 66 ^ { \circ } = - 1 + \tan 69 ^ { \circ } \tan 66 ^ { \circ }tan69∘+tan66∘=−1+tan69∘tan66∘

\tan 69 ^ { \circ } \tan 66 ^ { \circ } = 1 + \tan 66 ^ { \circ } + \tan 69 ^ { \circ }tan69∘tan66∘=1+tan66∘+tan69∘