Question

2a+cosa+tga, ctga=1

Answer 1

$ctg(a)=1\\a=arcctg1\\a=\frac{\pi}{4}$


$(sina)^2+\sqrt{2}cosa+tga=(sin\frac{\pi}{4})^2+\sqrt{2}*cos\frac{\pi}{4}+tg\frac{\pi}{4}=\\=(\frac{\sqrt{2}}{2})^2+\sqrt{2}*\frac{\sqrt{2}}{2}+1=\frac{1}{2}+1+1=2.5$

Answer 2

To solve this expression, we need to substitute ctga with 1 since ctga = 1.

So, we have:

2a + cosa + tga

= 2a + cosa + (1/cos(a))

Now, we need to simplify the expression by finding a common denominator for cosa and (1/cos(a)).

We know that:

cos(a)/cos(a) = 1

Therefore, we can multiply the first term by cos(a)/cos(a) and the second term by (1/cos(a))/(1/cos(a)):

2acos(a)/cos(a) + cosa(1/cos(a))/(1/cos(a)) + (1/cos(a))*cos(a)/cos(a)

= (2acos(a) + sin(a) + 1)/cos(a)

So, the simplified expression is:

(2acos(a) + sin(a) + 1)/cos(a)