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Sách Giáo Khoa
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Bùi Thị Vân
10 tháng 5 2017 lúc 9:44

a) \(sin6\alpha cot3\alpha cos6\alpha=2.sin3\alpha.cos3\alpha\dfrac{cos3\alpha}{sin3\alpha}-cos6\alpha\)
\(=2cos^23\alpha-\left(2cos^23\alpha-1\right)=1\) (Không phụ thuộc vào x).

Bùi Thị Vân
10 tháng 5 2017 lúc 9:56

b) \(\left[tan\left(90^o-\alpha\right)-cot\left(90^o+\alpha\right)\right]^2\)\(-\left[cot\left(180^o+\alpha\right)+cot\left(270^o+\alpha\right)\right]^2\)
\(=\left[cot\alpha+cot\left(90^o-\alpha\right)\right]^2\)\(-\left[cot\alpha+cot\left(90^o+\alpha\right)\right]^2\)
\(=\left[cot\alpha+tan\alpha\right]^2-\left[cot\alpha-tan\alpha\right]^2\)
\(=4tan\alpha cot\alpha=4\). (Không phụ thuộc vào \(\alpha\)).

Bùi Thị Vân
10 tháng 5 2017 lúc 10:05

c) \(\left(tan\alpha-tan\beta\right)cot\left(\alpha-\beta\right)-tan\alpha tan\beta\)
\(=\left(\dfrac{sin\alpha}{cos\alpha}-\dfrac{sin\beta}{cos\beta}\right).\dfrac{cos\left(\alpha-\beta\right)}{sin\left(\alpha-\beta\right)}-tan\alpha tan\beta\)
\(=\left(\dfrac{sin\alpha cos\beta-cos\alpha sin\beta}{cos\alpha cos\beta}\right).\dfrac{cos\left(\alpha-\beta\right)}{sin\left(\alpha-\beta\right)}\)\(-\dfrac{sin\alpha sin\beta}{cos\alpha cos\beta}\)
\(=\dfrac{sin\left(\alpha-\beta\right)}{cos\alpha cos\beta}.\dfrac{cos\left(\alpha-\beta\right)}{sin\left(\alpha-\beta\right)}-\dfrac{sin\alpha sin\beta}{cos\alpha cos\beta}\)
\(=\dfrac{cos\left(\alpha-\beta\right)}{cos\alpha cos\beta}-\dfrac{sin\alpha sin\beta}{cos\alpha cos\beta}\)
\(=\dfrac{cos\alpha cos\beta+sin\alpha sin\beta-sin\alpha sin\beta}{cos\alpha cos\beta}=\dfrac{cos\alpha cos\beta}{cos\alpha cos\beta}=1\).

Sách Giáo Khoa
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Bùi Thị Vân
11 tháng 5 2017 lúc 10:22

a) \(tan3\alpha-tan2\alpha-tan\alpha=\left(tan3\alpha-tan\alpha\right)-tan2\alpha\)
\(=\left(\dfrac{sin3\alpha}{cos3\alpha}-\dfrac{sin\alpha}{cos\alpha}\right)-\dfrac{sin2\alpha}{cos2\alpha}\)\(=\dfrac{sin3\alpha cos\alpha-cos3\alpha sin\alpha}{cos3\alpha cos\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\)
\(=\dfrac{sin2\alpha}{cos3\alpha cos\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\)
\(=sin2\alpha.\left(\dfrac{1}{cos3\alpha cos\alpha}-\dfrac{1}{cos2\alpha}\right)\)
\(=sin2\alpha.\dfrac{cos2\alpha-cos3\alpha cos\alpha}{cos3\alpha cos\alpha cos2\alpha}\)
\(=sin2\alpha.\dfrac{cos2\alpha-\dfrac{1}{2}\left(cos4\alpha+cos2\alpha\right)}{cos3\alpha cos2\alpha cos\alpha}\)
\(=sin2\alpha.\dfrac{cos2\alpha-cos4\alpha}{2cos3\alpha cos2\alpha cos\alpha}\)
\(=\dfrac{sin2\alpha.2sin3\alpha.sin\alpha}{2cos3\alpha cos2\alpha cos\alpha}\)
\(=tan3\alpha tan2\alpha tan\alpha\) (Đpcm).

Bùi Thị Vân
11 tháng 5 2017 lúc 10:30

b) \(\dfrac{4tan\alpha\left(1-tan^2\alpha\right)}{\left(1+tan^2\right)^2}=4tan\alpha\left(1-tan^2\alpha\right):\left(\dfrac{1}{cos^2\alpha}\right)^2\)
\(=4tan\alpha\left(1-tan^2\alpha\right)cos^4\alpha\)
\(=4\dfrac{sin\alpha}{cos\alpha}\left(1-\dfrac{sin^2\alpha}{cos^2\alpha}\right)cos^4\alpha\)
\(=4sin\alpha\left(cos^2\alpha-sin^2\alpha\right)cos\alpha\)
\(=4sin\alpha cos\alpha.cos2\alpha\)
\(=2.sin2\alpha.cos2\alpha=sin4\alpha\) (Đpcm).

Bùi Thị Vân
11 tháng 5 2017 lúc 10:35

c) \(\dfrac{1+tan^4\alpha}{tan^2\alpha+cot\alpha}=\left(1+tan^4\alpha\right):\left(tan^2\alpha+cot^2\alpha\right)\)
\(=\left(1+\dfrac{sin^4\alpha}{cos^4\alpha}\right):\left(\dfrac{sin^2\alpha}{cos^2\alpha}+\dfrac{cos^2\alpha}{sin^2\alpha}\right)\)
\(=\dfrac{sin^4\alpha+cos^4\alpha}{cos^4\alpha}:\dfrac{sin^4\alpha+cos^4\alpha}{cos^2\alpha sin^2\alpha}\)
\(=\dfrac{sin^2\alpha}{cos^2\alpha}=tan^2\alpha\) (Đpcm).

Nguyễn Sinh Hùng
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Akai Haruma
26 tháng 7 2021 lúc 14:47

Lời giải:
a.

$\tan a+\cot a=2\Leftrightarrow \tan a+\frac{1}{\tan a}=2$

$\Leftrightarrow \frac{\tan ^2a+1}{\tan a}=2$

$\Leftrightarrow \tan ^2a-2\tan a+1=0$

$\Leftrightarrow (\tan a-1)^2=0\Rightarrow \tan a=1$

$\cot a=\frac{1}{\tan a}=1$

$1=\tan a=\frac{\cos a}{\sin a}\Rightarrow \cos a=\sin a$

Mà $\cos ^2a+\sin ^2a=1$

$\Rightarrow \cos a=\sin a=\pm \frac{1}{\sqrt{2}}$

b.

Vì $\sin a=\cos a=\pm \frac{1}{\sqrt{2}}$

$\Rightarrow \sin a\cos a=\frac{1}{2}$

$E=\frac{\sin a.\cos a}{\tan ^2a+\cot ^2a}=\frac{\frac{1}{2}}{1+1}=\frac{1}{4}$

huy tạ
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huy tạ
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Họ Và Tên
21 tháng 10 2021 lúc 22:36

A

Nguyễn Lê Phước Thịnh
21 tháng 10 2021 lúc 22:37

Chọn A

Nguyễn Thùy Dương
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Akai Haruma
1 tháng 10 2018 lúc 23:28

a)

\(\sin ^4a-\cos ^4a+1=(\sin ^2a-\cos ^2a)(\sin ^2a+\cos^2a)+1\)

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

\(=2\sin ^2a\)

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

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

\(\Rightarrow \frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{\frac{\cos ^2a}{\sin ^2a}}=\sin ^2a\)

c)

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

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

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

Akai Haruma
1 tháng 10 2018 lúc 23:37

d)

\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}\) \(=\frac{\sin ^2a(1-\frac{1}{\cos ^2a})}{\cos ^2a(1-\frac{1}{\sin ^2a})}\)

\(=\frac{\sin ^2a.\frac{\cos ^2a-1}{\cos ^2a}}{\cos ^2a.\frac{\sin ^2a-1}{\sin ^2a}}\) \(=\frac{\sin ^2a.\frac{-\sin ^2a}{\cos ^2a}}{\cos ^2a.\frac{-\cos ^2a}{\sin ^2a}}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)

f)

\(\frac{(\sin a+\cos a)^2-1}{\cot a-\sin a\cos a}=\frac{\sin ^2a+\cos ^2a+2\sin a\cos a-1}{\frac{\cos a}{\sin a}-\sin a\cos a}\)

\(=\sin a.\frac{1+2\sin a\cos a-1}{\cos a-\cos a\sin ^2a}\)

\(=\sin a. \frac{2\sin a\cos a}{\cos a(1-\sin ^2a)}=\sin a. \frac{2\sin a\cos a}{\cos a. \cos^2 a}=\frac{2\sin ^2a}{\cos ^2a}=2\tan ^2a\)

Akai Haruma
1 tháng 10 2018 lúc 23:38

e)

\((1+\cot a)\sin ^3a+(1+\tan a)\cos ^3a\)

\(=(\sin ^3a+\cos ^3a)+\cot a.\sin ^3a+\tan a.\cos^3a\)

\(=(\sin a+\cos a)(\sin ^2a-\sin a\cos a+\cos ^2a)+\frac{\cos a}{\sin a}.\sin ^3a+\frac{\sin a}{\cos a}.\cos ^3a\)

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

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

\(=\sin a+\cos a\)

Nguyễn Sinh Hùng
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Nguyễn Việt Lâm
28 tháng 4 2021 lúc 19:13

\(P=\dfrac{\dfrac{sina}{cosa}+\dfrac{cosa}{sina}}{\dfrac{sina}{cosa}-\dfrac{3cosa}{sina}}=\dfrac{sin^2a+cos^2a}{sin^2a-3cos^2a}=\dfrac{1}{sin^2a-3\left(1-sin^2a\right)}=\dfrac{1}{4sin^2a-3}=\dfrac{1}{4.\left(\dfrac{1}{3}\right)^2-3}=...\)

Thầy Tùng Dương
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Cao Thị Kim Ngân
18 tháng 7 2022 lúc 10:42

a) Ta có A=\dfrac{\tan \alpha+3 \dfrac{1}{\tan \alpha}}{\tan \alpha+\dfrac{1}{\tan \alpha}}=\dfrac{\tan ^{2} \alpha+3}{\tan ^{2} \alpha+1}=\dfrac{\dfrac{1}{\cos ^{2} \alpha}+2}{\dfrac{1}{\cos ^{2} \alpha}}=1+2 \cos ^{2} \alpha Suy ra A=1+2 \cdot \dfrac{9}{16}=\dfrac{17}{8}.

b) B=\dfrac{\dfrac{\sin \alpha}{\cos ^{3} \alpha}-\dfrac{\cos \alpha}{\cos ^{3} \alpha}}{\dfrac{\sin ^{3} \alpha}{\cos ^{3} \alpha}+\dfrac{3 \cos ^{3} \alpha}{\cos ^{3} \alpha}+\dfrac{2 \sin \alpha}{\cos ^{3} \alpha}}=\dfrac{\tan \alpha\left(\tan ^{2} \alpha+1\right)-\left(\tan ^{2} \alpha+1\right)}{\tan ^{3} \alpha+3+2 \tan \alpha\left(\tan ^{2} \alpha+1\right)}.

Suy ra B=\dfrac{\sqrt{2}(2+1)-(2+1)}{2 \sqrt{2}+3+2 \sqrt{2}(2+1)}=\dfrac{3(\sqrt{2}-1)}{3+8 \sqrt{2}}.

Bảo Nam Phan
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Nguyễn Lê Phước Thịnh
13 tháng 7 2023 lúc 14:15

sin a=1/4

=>sin^2a=1/16

=>cos^2a=15/16

\(B=\dfrac{3\cdot\dfrac{cosa}{sina}-\dfrac{sina}{cosa}}{2\cdot\dfrac{sina}{cosa}+\dfrac{cosa}{sina}}\)

\(=\dfrac{3\cdot cosa^2a-sin^2a}{sina\cdot cosa}:\dfrac{2\cdot sin^2a+cos^2a}{sina\cdot cosa}\)

\(=\dfrac{3\cdot cos^2a-sin^2a}{2\cdot sin^2a+cos^2a}\)

\(=\dfrac{3\cdot\dfrac{15}{16}-\dfrac{1}{16}}{2\cdot\dfrac{1}{16}+\dfrac{15}{16}}=\dfrac{44}{17}\)