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a) 1 + tan22 a =1 +(\(\dfrac{sina}{cosa}\))2 =\(\dfrac{sina+cosa}{cos^2a}\)=\(\dfrac{1}{cos^2a}\)
b) 1 + cot2 a= 1 +(\(\dfrac{cosa}{sina}\))2 = \(\dfrac{cosa+sina}{sin^2a}\)=\(\dfrac{1}{sin^2a}\)
c) tan2 a (2 sin2a + 3 cos2 a - 2)
=tan2 a[cos2 a +2 (\(sina^2+cos^2a\))-2 ]
=\(\dfrac{sin^2a}{cos^2a}\)×\(cos^2a=sin^2a\)
b: \(1+cot^2a=1+\left(\dfrac{cosa}{sina}\right)^2=\dfrac{1}{sin^2a}\)
c: \(=tan^2a\left[2\left(1-cos^2a\right)+3cos^2a-2\right]\)
\(=tan^2a\left[cos^2a\right]\)
\(=\dfrac{sin^2a}{cos^2a}\cdot cos^2a=sin^2a\)
\(\left(\sqrt{\dfrac{1+sin\alpha}{1-sin\alpha}}+\sqrt{\dfrac{1-sin\alpha}{1+sin\alpha}}\right).\dfrac{1}{\sqrt{1+tan^2\alpha}}\)
\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{\left(1-sin\alpha\right)\left(1+sin\alpha\right)}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{\left(1+sin\alpha\right)\left(1-sin\alpha\right)}}\right).\dfrac{1}{\sqrt{1+\left(\dfrac{sin\alpha}{cos\alpha}\right)^2}}\)
\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{1-sin^2\alpha}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{1-sin^2\alpha}}\right).\dfrac{1}{\sqrt{\dfrac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}}}\)
\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{cos^2\alpha}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{cos^2\alpha}}\right).\dfrac{1}{\sqrt{\dfrac{1}{cos^2\alpha}}}\)
\(=\left(\dfrac{1+sin\alpha}{cos\alpha}+\dfrac{1-sin\alpha}{cos\alpha}\right).\dfrac{1}{\dfrac{1}{cos\alpha}}=\dfrac{2}{cos\alpha}.cos\alpha=2\)
Đề bài ko chính xác, biểu thức này không rút gọn được (có thể coi việc biến đổi khả dĩ duy nhất \(1+2sina.cosa=\left(sina+cosa\right)^2\) không phải là hành động rút gọn)
chỉnh lại đề 1 chút: \(A=\dfrac{1+2sin\alpha.cos\alpha}{cos^2\alpha-sin^2\alpha}=\dfrac{cos^2\alpha+sin^2\alpha+2sin\alpha.cos\alpha}{\left(cos\alpha-sin\alpha\right)\left(cos\alpha+sin\alpha\right)}\)
\(=\dfrac{\left(cos\alpha+sin\alpha\right)^2}{\left(cos\alpha-sin\alpha\right)\left(cos\alpha+sin\alpha\right)}=\dfrac{cos\alpha+sin\alpha}{cos\alpha-sin\alpha}\)
a. \(\dfrac{1+2sin\alpha cos\alpha}{cos^2\alpha-sin^2\alpha}=\dfrac{sin^2\alpha+2sin\alpha cos\alpha+cos^2}{\left(cos\alpha-sin\alpha\right)\left(cos\alpha+sin\alpha\right)}=\dfrac{\left(sin\alpha+cos\alpha\right)^2}{\left(cos\alpha-sin\alpha\right)\left(cos\alpha+sin\alpha\right)}=\dfrac{sin\alpha+cos\alpha}{cos\alpha-sin\alpha}\)
a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)
\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)
\(=\dfrac{\left(a+1\right)\left(a-1\right)}{1-a}\)
=-a-1
b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)
\(=\left|3a-5\right|-2a+4\)
\(=5-3a-2a+4\)
=9-5a
c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)
\(=4a-3-\left|2a-1\right|\)
\(=4a-3-2a+1\)
\(=2a-2\)
d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)
\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)
\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)
\(=-a^2\)
a, \(\tan^2\alpha\left(2\cos^2\alpha+\sin^2\alpha-1\right)\)
\(=\tan^2\alpha\left(\cos^2\alpha+\cos^2\alpha+\sin^2\alpha-1\right)\)
\(=\tan^2\alpha\left(\cos^2\alpha+1-1\right)\)
\(=\tan^2\alpha.\cos^2\alpha=1\)
b, \(\sin\alpha-\sin\alpha.\cos^2\alpha\)
\(=\sin\alpha\left(1-\cos^2\alpha\right)\)
\(=\sin\alpha.\sin^2\alpha\)
bn ơi lm j có công thức \(\tan^2a\times\cos^2a=1\) đâu
\(=\dfrac{1-\sin a+1+\sin a}{1-\sin^2a}-2\tan^2a\)
\(=\dfrac{2}{\cos^2a}-2\cdot\dfrac{\sin^2a}{\cos^2a}\)
\(=\dfrac{2\left(1-\sin^2a\right)}{\cos^2a}=2\)