\(\left(\sqrt{\dfrac{1+sin\alpha}{1-sin\alpha}}+\sqrt{\dfrac{1-sin\alpha}{1+sin\alpha}}\right).\dfrac{1}{\sqrt{1+tan^2\alpha}}\)

\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{\left(1-sin\alpha\right)\left(1+sin\alpha\right)}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{\left(1+sin\alpha\right)\left(1-sin\alpha\right)}}\right).\dfrac{1}{\sqrt{1+\left(\dfrac{sin\alpha}{cos\alpha}\right)^2}}\)

\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{1-sin^2\alpha}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{1-sin^2\alpha}}\right).\dfrac{1}{\sqrt{\dfrac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}}}\)

\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{cos^2\alpha}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{cos^2\alpha}}\right).\dfrac{1}{\sqrt{\dfrac{1}{cos^2\alpha}}}\)

\(=\left(\dfrac{1+sin\alpha}{cos\alpha}+\dfrac{1-sin\alpha}{cos\alpha}\right).\dfrac{1}{\dfrac{1}{cos\alpha}}=\dfrac{2}{cos\alpha}.cos\alpha=2\)

a) Thu gọn B = -8;         b) Thu gọn C = 2018.

a/ \(A=\frac{cot^2a-cos^2a}{cot^2a}-\frac{sina.cosa}{cota}\)

\(=\frac{\frac{cos^2a}{sin^2a}-cos^2a}{\frac{cos^2a}{sin^2a}}-\frac{sina.cosa}{\frac{cosa}{sina}}\)

\(=\left(1-sin^2a\right)-sin^2a=1\)

b/ \(B=\left(cosa-sina\right)^2+\left(cosa+sina\right)^2+cos^4a-sin^4a-2cos^2a\)

\(=cos^2a-2cosa.sina+sin^2a+cos^2a+2cosa.sina+sin^2a+\left(cos^2a+sin^2a\right)\left(cos^2a-sin^2a\right)-2cos^2a\)

\(=2+\left(cos^2a-sin^2a\right)-2cos^2a\)

\(=2-sin^2a-cos^2a=2-1=1\)

c/ \(C=sin^6x+cos^6x+3sin^2x.cos^2x\)

\(=\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cos^2x+cos^4x\right)+3sin^2x.cos^2x\)

\(=sin^4x-sin^2x.cos^2x+cos^4x+3sin^2x.cos^2x\)

\(=sin^4x+cos^4x+2sin^2x.cos^2x\)

\(=\left(sin^2x+cos^2x\right)^2=1\)

b) \(\frac{\sin25+\cos70}{\sin20+\cos65}\)

xét tam giác vuông có :  sin a= cos b => cos 70 = sin (90 -70)  <=> cos 70 = sin 20

                                    cos 65 =sin 25

<=> \(\frac{\sin25+\cos70}{\sin20+\cos65}\)​​

=\(\frac{\sin25+\sin20}{\sin20+\sin25}=1\)

 \(\frac{2\cos^2\cdot a-1}{\sin a+\cos a}=\frac{2\cos^2a-\left(\sin^2+\cos^2\right)}{\sin a+\cos a}\)

vì \(\sin^2a+\cos^2a=1\)

=\(\frac{\cos^2a-\sin^2a}{\sin a+\cos a}=\frac{\left(\cos a-\sin a\right)\left(\cos a+\sin a\right)}{\sin a+\cos a}\)

=\(\cos a-\sin a\)

a/. ĐKXĐ : (x-1)(x+1) # 0 => x # 1 hay x # -1

b/. \(B=\left[\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{3.2}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}\right].\frac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(B=\frac{x^2+2x+1+6-x^2-4x-3}{2\left(x-1\right)\left(x+1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)

\(B=\frac{2\left(4-2x\right)}{5}\)

Em xem lại đè nhé. Đề như vậy thì sẽ ko rút gọn đc hết x trên tử. nên B vẫn phụ thuộc vào biến x. 

chao cac bạn và a chi nếu đề sửa lai vây thi minh làm thế nào ( x+1/2x-2 + 3/x^2+1 - x+3/2x+1 )* (4x^2 -1)/5

\(\frac{x+3}{2x+2}\)

sẽ ra đó bạn

\(B=cos^2a+sin^2a\left(cos^2a+sin^2a\right)=cos^2a+sin^2a=1\)

\(C=\frac{1-sina+1+sina}{\left(1+sina\right)\left(1-sina\right)}-2tan^2a=\frac{2}{1-sin^2a}-2tan^2a\)

\(=\frac{2}{cos^2a}-\frac{2sin^2a}{cos^2a}=\frac{2\left(1-sin^2a\right)}{cos^2a}=\frac{2cos^2a}{cos^2a}=2\)

\(\sin^4\alpha+\sin^2\alpha.\cos^2\alpha+\cos^2\alpha=\)\(\sin^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)+\cos^2\alpha=\sin^2\alpha+\cos^2\alpha=1\)

\(\frac{1}{1+\sin\alpha}+\frac{1}{1-\sin\alpha}-2\tan^2\alpha=\frac{1-\sin\alpha+1+\sin\alpha}{1-\sin^2\alpha}-\frac{2\sin^2\alpha}{\cos^2\alpha}=\)

\(\frac{2}{1-\sin^2\alpha}-\frac{2\sin^2\alpha}{\cos^2\alpha}=2\left(\frac{1}{\cos^2\alpha}-\frac{\sin^2\alpha}{\cos^2\alpha}\right)=2\)

chúng không phụ thuộc vào số đo góc\(\alpha\)

I don't now

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