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AH
Akai Haruma
Giáo viên
24 tháng 6 2020

Bài 2:

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

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

Ta có đpcm.

AH
Akai Haruma
Giáo viên
24 tháng 6 2020

1.

$0< a< 90^0\Rightarrow `1>\sin a, \cos a>0$

Do đó:

$\sin a-\tan a=\sin a-\frac{\sin a}{\cos a}=\frac{\sin a(\cos a-1)}{\cos a}<0$

$\Rightarrow \sin a< \tan a$

(đpcm)

$\cos a-\cot a=\cos a-\frac{\cos a}{\sin a}=\frac{\cos a(\sin a-1)}{\sin a}<0$

$\Rightarrow \cos a< \cot a$ (đpcm)

 

15 tháng 7 2018

b,ta có :\(\frac{sin^2a-cos^2a\left(1-cos^2a\right)}{cos^2a-sin^2a\left(1-sin^2a\right)}=\frac{sin^4a}{cos^4a}\)

=>\(\frac{sin^2a-sin^2a.cos^2a}{cos^2a-sin^2a.cos^2a}=\frac{sin^4a}{cos^4a}\)

=>\(\frac{sin^2a\left(1-cos^2a\right)}{cos^2a\left(1-sin^2a\right)}=\frac{sin^4a}{cos^4a}\)

=>\(\frac{sin^4a}{cos^4a}=\frac{sin^4a}{cos^4a}\)luon dung => dpcm

AH
Akai Haruma
Giáo viên
23 tháng 9 2018

Lời giải:

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

b)

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

c) Đề bài sai.

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

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

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

d)

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

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

e) ĐK tồn tại tan là $\cos x\neq 0$

\(\tan a=\frac{\sin a}{\cos a}\Rightarrow \sin a=\tan a\cos a\)

Ta có:

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

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

\(=\frac{\tan a\cos a-\cos a}{\tan a\cos a+\cos a}=\frac{\cos a(\tan a-1)}{\cos a(\tan a+1)}\)\(=\frac{\tan a-1}{\tan a+1}\) (đpcm)

27 tháng 9 2018

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: \(=tan^2a\left[2\left(1-cos^2a\right)+3cos^2a-2\right]\)

\(=tan^2a\left[cos^2a\right]\)

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

27 tháng 6 2017

a.Ta có \(\tan\alpha.\cot\alpha=1\Rightarrow\tan\alpha=\frac{1}{\cot\alpha}\)

\(\Rightarrow\frac{1}{\cot\alpha}+\cot\alpha=2\Rightarrow\cot^2\alpha-2\cot\alpha+1=0\)

\(\cot\alpha=1\Rightarrow\alpha=45^0\)

b.Ta có \(\sin^2\alpha+\cos^2\alpha=1\Rightarrow\cos^2\alpha=1-\sin^2\alpha\)

\(\Rightarrow7.\sin^2\alpha+5\left(1-\sin^2\alpha\right)=\frac{13}{2}\)\(\Leftrightarrow\sin^2\alpha=\frac{3}{4}\Leftrightarrow\orbr{\begin{cases}sin\alpha=\frac{\sqrt{3}}{2}\\sin\alpha=\frac{-\sqrt{3}}{2}\end{cases}}\)

\(\Rightarrow\alpha=60^0\)

16 tháng 7 2021

B A C a

Xét ΔBAC vuông tại B có a = ^A ta có :

a) \(\frac{\sin\alpha}{\cos\alpha}=\frac{\sin A}{\cos A}=\frac{\frac{BC}{AB}}{\frac{AB}{AC}}=\frac{BC}{AB}\cdot\frac{AC}{AB}=\frac{BC}{AB}=\tan A=\tan\alpha\left(đpcm\right)\)

b) \(\frac{\cos\alpha}{\sin\alpha}=\frac{\cos A}{\sin A}=\frac{\frac{AB}{AC}}{\frac{BC}{AC}}=\frac{AB}{AC}\cdot\frac{AC}{BC}=\frac{AB}{BC}=\cot A=\cot\alpha\left(đpcm\right)\)

c) \(\tan\alpha\cdot\cot\alpha=\tan A\cdot\cot A=\frac{BC}{AB}\cdot\frac{AB}{BC}=1\left(đpcm\right)\)

d) \(\sin^2\alpha+\cos^2\alpha=\sin^2A+\cos^2A=\frac{BC^2}{AC^2}+\frac{AB^2}{AC^2}=\frac{AB^2+BC^2}{AC^2}=1\left(đpcm\right)\)

e) \(\frac{1}{\cos^2\alpha}=\frac{1}{\cos^2A}=\frac{1}{\frac{AB^2}{AC^2}}=\frac{AC^2}{AB^2};1+\tan^2\alpha=1+\tan^2A=1+\frac{BC^2}{AB^2}=\frac{AB^2+BC^2}{AB^2}=\frac{AC^2}{AB^2}\)

\(\Rightarrow1+\tan^2\alpha=\frac{1}{\cos^2\alpha}\left(đpcm\right)\)

f) \(\frac{1}{\sin^2\alpha}=\frac{1}{\sin^2A}=\frac{1}{\frac{BC^2}{AC^2}}=\frac{AC^2}{BC^2};1+\cot^2\alpha=1+\cot^2A=1+\frac{AB^2}{BC^2}=\frac{BC^2+AB^2}{BC^2}=\frac{AC^2}{BC^2}\)

\(\Rightarrow1+\cot^2\alpha=\frac{1}{\sin^2\alpha}\left(đpcm\right)\)

7 tháng 8 2017

~ ~ ~ Áp dụng đẳng thức \(\left(a+b\right)^2+\left(a-b\right)^2=2\left(a^2+b^2\right)\) ~ ~ ~

a)

\(\left(\sin\alpha+\cos\alpha\right)^2-2\sin\alpha\cos\alpha-1\)

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

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

= 0

b)

\(\left(\sin\alpha-\cos\alpha\right)^2+2\sin\alpha\cos\alpha+1\)

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

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

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

= 2

c)

\(\left(\sin\alpha+\cos\alpha\right)^2+\left(\sin\alpha-\cos\alpha\right)^2+2\)

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

= 4

d)

\(\sin^2\alpha\cot^2\alpha+\cos^2\alpha\tan^2\alpha\)

\(=\left(\sin\times\dfrac{\cos}{\sin}\right)^2+\left(\cos\times\dfrac{\sin}{\cos}\right)^2\)

= 1

21 tháng 10 2019

sin a = \(\frac{2}{3}\) => \(\sin^2a\) = \(\frac{4}{9}\) => \(\cos^2a\) = \(1-\frac{4}{9}=\frac{5}{9}\) => \(\cos a\) = \(\frac{\sqrt{5}}{3}\)

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

P = \(\left(\frac{2}{3}:\frac{\sqrt{5}}{3}\right)^2-2\left(\frac{\sqrt{5}}{3}:\frac{2}{3}\right)^2\)

P = \(\left(\frac{2\sqrt{5}}{5}\right)^2-2\left(\frac{\sqrt{5}}{2}\right)^2\)

P = \(\frac{4}{5}-\frac{5}{2}\)

P = \(\frac{-17}{10}\)