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

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вопрос

To expand the expression (1+sina)(tga+ctga)(1-sina), we can first expand the first two terms using the distributive property:

(1+sina)(tga+ctga) = tga + ctga + sin(tga) + ctsina

Next, we can multiply the above result by (1-sina) using the distributive property again:

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

Expanding further:

tga - tgasina + ctga - ctgasina + sin(tga) - sin(tga)sina + ctsina - ctsinasina

Finally, simplify the expression:

tga - tgasina + ctga - ctgasina + sin(tga) - sin(tga)sina + ctsina - ctsinasina

= tga + ctga + sin(tga) + ctsina - (tga^2)sina - ctga^2sina + sin(tga) - sin(tga)sina

Therefore, the expanded expression is tga + ctga + sin(tga) + ctsina - (tga^2)sina - ctga^2sina + sin(tga) - sin(tga)sina.