Question

tg(9) × tg(81) + tg(27) × tg63

Answer 1

Ответ:

Пошаговое объяснение:

tg9*tg81+tg27*tg63=tg(90-81)*tg81+tg(90-63)*tg63=

=ctg81*tg81+ctg63*tg63=1+1=2

Answer 2

Let's use the identity:
$$\tan(A+B) = \frac{\tan A + \tan B}{1-\tan A \tan B}$$
For $A=9$ and $B=81$, we have:
$$\tan(9+81) = \frac{\tan 9 + \tan 81}{1-\tan 9 \tan 81}$$
Rearranging, we get:
$$\tan 9 \tan 81 = \frac{\tan 9 + \tan 81}{\tan(9+81)} - 1$$
We can use the same identity for $A=27$ and $B=63$ to get:
$$\tan 27 \tan 63 = \frac{\tan 27 + \tan 63}{\tan(27+63)} - 1$$
Now we just need to substitute the values into the expression given:
\begin{align*}
\tan 9 \tan 81 + \tan 27 \tan 63 &= \left(\frac{\tan 9 + \tan 81}{\tan(9+81)} - 1\right) \\
&\quad\quad+ \left(\frac{\tan 27 + \tan 63}{\tan(27+63)} - 1\right)\\
&= \frac{\tan 9 + \tan 81}{\tan(9+81)} + \frac{\tan 27 + \tan 63}{\tan(27+63)} - 2
\end{align*}
Now we just need to simplify the two fractions in the expression:
\begin{align*}
\frac{\tan 9 + \tan 81}{\tan(9+81)} &= \frac{\tan 9 + \tan(90-9)}{\tan(90)} \\
&= \frac{\tan 9 + \cot 9}{1} \\
&= \tan 9 \sec 9 + 1 \\
&= \frac{\sin 9}{\cos 9} \cdot \frac{1}{\cos 9} + 1\\
&= \frac{\sin 9 + \cos^2 9}{\cos^2 9}
\end{align*}
Similarly,
\begin{align*}
\frac{\tan 27 + \tan 63}{\tan(27+63)} &= \frac{\tan 27 + \tan(90-27)}{\tan(90)} \\
&= \frac{\tan 27 + \cot 27}{1} \\
&= \tan 27 \sec 27 + 1 \\
&= \frac{\sin 27}{\cos 27} \cdot \frac{1}{\cos 27} + 1\\
&= \frac{\sin 27 + \cos^2 27}{\cos^2 27}
\end{align*}
Substituting these into the expression we found earlier, we get:
\begin{align*}
\tan 9 \tan 81 + \tan 27 \tan 63 &= \frac{\sin 9 + \cos^2 9}{\cos^2 9} + \frac{\sin 27 + \cos^2 27}{\cos^2 27} - 2 \\
&= \frac{\sin 9}{\cos^2 9} + \frac{\sin 27}{\cos^2 27} \\
&= \boxed{\csc^2 9 + \csc^2 27}
\end{align*}