Sin105° +Sin75° =? Найдите sqrt(tga) + sqrt(ctga) если tga+ctga=a(a>0)

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To find sin105° and sin75°, we must use the following identities:

sin(A + B) = sinA cosB + cosA sinB
sin(A - B) = sinA cosB - cosA sinB

Therefore:
sin(105°) = sin(45° + 60°) = sin45° cos60° + cos45° sin60°
sin(105°) = (√2/2)(1/2) + (√2/2)(√3/2) = √2/4 + √6/4 = (√2 + √6)/4

sin(75°) = sin(45° + 30°) = sin45° cos30° + cos45° sin30°
sin(75°) = (√2/2)(√3/2) + (√2/2)(1/2) = √6/4 + √2/4 = (√6 + √2)/4

Therefore, sin105° + sin75° = (√2 + √6)/4 + (√6 + √2)/4 = 2√2/4 = √2

Given tga + ctga = a (a > 0), we can rewrite as:

tan(a) + cot(a) = a

Now, we want to find sqrt(tan(a)) + sqrt(cot(a))

Let x = sqrt(tan(a)) and y = sqrt(cot(a))

Since tan(a) = 1/cot(a), we have:

x^2 + y^2 = tan(a) + cot(a) = a

Therefore, sqrt(tan(a)) + sqrt(cot(a)) = x + y = sqrt(x^2 + y^2) = sqrt(a)