For α∈(0∘;90∘) simplify sin2α+tan2α+sin2α⋅cos2α+cos4α My try: sin2α+tan2α+sin2α⋅cos2α+cos4α=sin2α+sin2αcos2α+sin2α⋅cos2α+cos4α =sin2α⋅cos2α+sin2+sin2α⋅cos4α+cos6αcos2α

Question

For α∈(0∘;90∘) simplify sin2α+tan2α+sin2α⋅cos2α+cos4α  My try: sin2α+tan2α+sin2α⋅cos2α+cos4α=sin2α+sin2αcos2α+sin2α⋅cos2α+cos4α  =sin2α⋅cos2α+sin2+sin2α⋅cos4α+cos6αcos2α

poleglit3 · Accepted Answer

Express everything in terms of cos⁡α=c  (1−c2)+(1−c2)c2+(1−c2)c2+c4  =c2−c4+1−c2+c4−c6+c6c2=1c2

RizerMix · Accepted Answer

Youre

user_27qwe · Accepted Answer

Instead of dividing by cos2⁡α, you could do the following, sin2⁡α+tan2⁡α+sin2⁡α+sin2⁡αcos2⁡α+cos4⁡α =sin2⁡α+tan2⁡α+sin2⁡α+cos2⁡α(sin2⁡α+cos2⁡α) =sin2⁡α+tan2⁡α+cos2⁡α =1+tan2⁡α =sec2⁡α

For α∈(0∘;90∘) simplify sin2α+tan2α+sin2α⋅cos2α+cos4α My try: sin2α+tan2α+sin2α⋅cos2α+cos4α=sin2α+sin2αcos2α+sin2α⋅cos2α+cos4α =sin2α⋅cos2α+sin2+sin2α⋅cos4α+cos6αcos2α

Answered question

Answer & Explanation