Question

: tg 9° tg 81°+ tg 27º. tg 63°​

Answer 1

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

по формуле приведения:

$tg(90-x)=ctgx$

$ctg81*tg81+ctg63*tg63$

$ctgx=\frac{1}{tgx}$

$\frac{1}{tg81}*tg81+\frac{1}{tg63}*tg63=1+1=2$

Answer 2

It seems like you've provided a sequence of trigonometric expressions involving tangent values of various angles. Let's solve each of them step by step:

1. tg(9°): The tangent of 9 degrees is a trigonometric value. If you want the numerical value, you can calculate it using a calculator or software: tg(9°) ≈ 0.15838444.

2. tg(81°): The tangent of 81 degrees is another trigonometric value. Numerically, tg(81°) ≈ 5.67128182.

3. tg(27°): The tangent of 27 degrees is yet another trigonometric value. Numerically, tg(27°) ≈ 0.50952024.

4. tg(63°): The tangent of 63 degrees is the last trigonometric value in the sequence. Numerically, tg(63°) ≈ 1.10714872.

Now, if you want to perform operations on these values as indicated by your expression, you can plug in the numerical values and perform the calculations:

tg(81°) + tg(27°) = 5.67128182 + 0.50952024 ≈ 6.18080206

tg(63°) = 1.10714872

So, the final expression becomes: tg(81°) + tg(27°) + tg(63°) ≈ 6.18080206 + 1.10714872 ≈ 7.28795078