Question

(1+sina)(tga+ctga)(1-sina)=ctga

Answer 1

Объяснение:

Задание: доказать тождество $(1+sin\alpha )(tg\alpha+ctg\alpha)(1-sin\alpha ) = ctg\alpha$

Упростим левую часть:

$(1+sin\alpha )(1-sin\alpha )(\frac{sin\alpha }{cos\alpha } +\frac{cos\alpha }{sin\alpha } )=(1-sin^2\alpha )(\frac{sin^2\alpha +cos^2\alpha }{sin\alpha cos\alpha } )=\\cos^2\alpha \frac{1}{sin\alpha cos\alpha } =\frac{cos\alpha }{sin\alpha } =ctg\alpha$, ч.т.д.

Answer 2

Let's simplify the given expression step by step:

(1 + sin(a))(tg(a) + ctg(a))(1 - sin(a)) = ctg(a)

Step 1: Distribute (1 + sin(a)) to (tg(a) + ctg(a)): tg(a) + ctg(a) + sin(a) * tg(a) + sin(a) * ctg(a)

Step 2: Distribute (1 - sin(a)) to the result of Step 1: tg(a) + ctg(a) + sin(a) * tg(a) + sin(a) * ctg(a) - sin(a) * tg(a) - sin(a) * ctg(a)

Step 3: Simplify the expression: tg(a) + ctg(a) - sin(a) * tg(a) + sin(a) * ctg(a)

Step 4: Factor out common terms: tg(a) - sin(a) * tg(a) + ctg(a) + sin(a) * ctg(a)

Step 5: Combine like terms: tg(a) * (1 - sin(a)) + ctg(a) * (1 + sin(a))

Now, it seems that the expression you provided is not equal to ctg(a) as mentioned at the end. There might be a mistake in the original equation or some missing information. Please double-check the equation or provide any additional information to resolve the discrepancy.