Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, Sử dụng tích chéo:
Ta có:
+/ \(\cos\alpha.\cos\alpha=\cos^2\alpha\) (1)
+/ \(\left(1+\sin\alpha\right)\left(1-\sin\alpha\right)=1-\sin^2\alpha\)
Mà \(\sin^2\alpha+\cos^2\alpha=1\)
\(\Rightarrow1-\sin^2\alpha=\cos^2\alpha\)
hay \(\left(1+\sin\alpha\right)\left(1-\sin\alpha\right)=\cos^2\alpha\) (2)
Từ (1), (2)
\(\Rightarrow\)\(\cos\alpha.\cos\alpha=\)\(\left(1+\sin\alpha\right)\left(1-\sin\alpha\right)\)
\(\Rightarrow\)\(\dfrac{\cos\alpha}{1-\sin\alpha}=\dfrac{1+\sin\alpha}{\cos\alpha}\) (đpcm)
b/ xem lại đề
a) \(\dfrac{2sina+3cosa}{3sina-4cosa}=\dfrac{9}{5}\)
b) \(\dfrac{sina.cosa}{sin^2a-sina.cosa+cos^2a}=0\)
\(a.\dfrac{2\sin\alpha+3\cos\alpha}{3\sin\alpha-4\cos\alpha}=\dfrac{2\left(3cos\alpha\right)+3cos\alpha}{3\left(3cos\alpha\right)-4cos\alpha}=\dfrac{9cos\alpha}{5cos\alpha}=\dfrac{9}{5}\)
\(b.\dfrac{sin\alpha cos\alpha}{sin^2\alpha-sin\alpha cos\alpha+cos^2\alpha}=\dfrac{3cos^2\alpha}{9cos^2\alpha-3cos^2\alpha+cos^2\alpha}=\dfrac{3cos^2\alpha}{7cos^2\alpha}=\dfrac{3}{7}\)
Bài 2:
a: \(\sin a=\sqrt{1-\left(\dfrac{4}{5}\right)^2}=\dfrac{3}{5}\)
\(P=4\cdot\sin^2a-6\cdot\cos^2a\)
\(=4\cdot\dfrac{9}{25}-6\cdot\dfrac{16}{25}\)
\(=\dfrac{36-64}{25}=\dfrac{-28}{25}\)
b: \(A=\sin^6a+\cos^6a+3\cdot\sin^2a\cdot\cos^2a\)
\(=\left(\sin^2a+\cos^2a\right)^3-3\sin^2a\cdot\cos^2a\cdot\left(\sin^2a+\cos^2a\right)+3\cdot\sin^2a\cdot\cos^2a\)
\(=1-3\sin^2a\cdot\cos^2a+3\sin^2a\cdot\cos^2a\)
=1
a) ta có : \(A=tan1.tan2.tan3...tan89\)
\(=\left(tan1.tan89\right).\left(tan2.tan88\right).\left(tan3.tan87\right)...\left(tan44.tan46\right).tan45\)
\(=\left(tan1.tan\left(90-1\right)\right).\left(tan2.tan\left(90-2\right)\right).\left(tan3.tan\left(90-3\right)\right)...\left(tan44.tan\left(90-44\right)\right).tan45\)
\(=\left(tan1.cot1\right).\left(tan2.cot2\right).\left(tan3.cot3\right)...\left(tan44.cot44\right).tan45\) \(=tan45=1\)b) ta có \(B=\dfrac{sin\alpha+2cos\alpha}{3sin\alpha-4cos\alpha}=\dfrac{\dfrac{sin\alpha}{cos\alpha}+\dfrac{2cos\alpha}{cos\alpha}}{\dfrac{3sin\alpha}{cos\alpha}-\dfrac{4cos\alpha}{cos\alpha}}\)
\(=\dfrac{tan\alpha+2}{3tan\alpha-4}=\dfrac{\dfrac{1}{2}+2}{\dfrac{3}{2}-4}=-1\)
ta có \(D=\dfrac{2sin^2\alpha-3cos^2\alpha}{4cos^2\alpha-5sin^2\alpha}=\dfrac{\dfrac{2sin^2\alpha}{cos^2\alpha}-\dfrac{3cos^2\alpha}{cos^2\alpha}}{\dfrac{4cos^2\alpha}{cos^2\alpha}-\dfrac{5sin^2\alpha}{cos^2\alpha}}\)
\(=\dfrac{2tan^2\alpha-3}{4-5tan^2\alpha}=\dfrac{2\left(\dfrac{1}{2}\right)^2-3}{4-5\left(\dfrac{1}{2}\right)^2}=\dfrac{-10}{11}\)
a) Ta có : sin\(^2\)12o=cos278o=> sin212o+sin278o=1.
tương tự => A=3
b) tương tự câu (a) ta có: cos215o=sin275o ( do 15+75=90 nha bạn ) => cos215o+cos275o=1. Tương tự => B=0
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 $\cos x\neq 0$
Vì \(\tan a=\frac{\sin a}{\cos a}\Rightarrow \sin a=\tan a\cos a\)
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)
Lời giải:
a) Áp dụng công thức \(\sin ^2a+\cos ^2a=1\) thì:
\(P=3\sin ^2a+4\cos ^2a=3(\sin ^2a+\cos ^2a)+\cos ^2a\)
\(=3.1+(\frac{1}{3})^2=\frac{28}{9}\)
b)
\(\tan a=\frac{3}{4}\Rightarrow \cot a=\frac{1}{\tan a}=\frac{4}{3}\)
\(\frac{3}{4}=\tan a=\frac{\sin a}{\cos a}\Rightarrow \sin a=\frac{3}{4}\cos a\)
\(\Rightarrow \sin ^2a=\frac{9}{16}\cos ^2a\)
\(\Rightarrow \sin ^2a+\cos ^2a=\frac{25}{16}\cos ^2a\Rightarrow \frac{25}{16}\cos ^2a=1\)
\(\Rightarrow \cos ^2a=\frac{16}{25}\Rightarrow \cos a=\pm \frac{4}{5}\)
Nếu \(\Rightarrow \sin a=\pm \frac{3}{5}\) (theo thứ tự)
c)
\(\frac{1}{2}=\tan a=\frac{\sin a}{\cos a}\Rightarrow \sin a=\frac{\cos a}{2}\). Vì a góc nhọn nên \(\cos a\neq 0\)
Do đó:
\(\frac{\cos a-\sin a}{\cos a+\sin a}=\frac{\cos a-\frac{\cos a}{2}}{\cos a+\frac{\cos a}{2}}=\frac{\cos a(1-\frac{1}{2})}{\cos a(1+\frac{1}{2})}=\frac{1-\frac{1}{2}}{1+\frac{1}{2}}=\frac{1}{3}\)
a) \(1+tan^2\alpha=1+\left(\dfrac{sin\alpha}{cos\alpha}\right)^2=\dfrac{sin^2\alpha+cos^2\alpha}{cos^2\alpha}=\dfrac{1}{cos^2\alpha}\)
b) \(1+cot^2\alpha=1+\left(\dfrac{cos\alpha}{sin\alpha}\right)^2=\dfrac{cos^2\alpha+sin^2\alpha}{sin^2\alpha}=\dfrac{1}{sin^2\alpha}\)
c) \(tan^2\alpha\left(2sin^2\alpha+3cos^2\alpha-2\right)=tan^2\alpha\left[cos^2\alpha+2\left(sin^2\alpha+cos^2\alpha\right)-2\right]=\dfrac{sin^2\alpha}{cos^2\alpha}\times cos^2\alpha=sin^2\alpha\)
a)
\(1+tan^2\alpha=1+\left(\dfrac{sin\alpha}{cos\alpha}\right)^2=\dfrac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}=\dfrac{1}{cos^2\alpha}\)
b)\(1+cot^2\alpha=1+\left(\dfrac{cos\alpha}{sin\alpha}\right)^2=\dfrac{sin^2\alpha+cos^2\alpha}{sin^2\alpha}=\dfrac{1}{sin^2\alpha}\)
c) mình chưa rõ đề nha