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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/ \(A=\frac{cot^2a-cos^2a}{cot^2a}-\frac{sina.cosa}{cota}\)
\(=\frac{\frac{cos^2a}{sin^2a}-cos^2a}{\frac{cos^2a}{sin^2a}}-\frac{sina.cosa}{\frac{cosa}{sina}}\)
\(=\left(1-sin^2a\right)-sin^2a=1\)
b/ \(B=\left(cosa-sina\right)^2+\left(cosa+sina\right)^2+cos^4a-sin^4a-2cos^2a\)
\(=cos^2a-2cosa.sina+sin^2a+cos^2a+2cosa.sina+sin^2a+\left(cos^2a+sin^2a\right)\left(cos^2a-sin^2a\right)-2cos^2a\)
\(=2+\left(cos^2a-sin^2a\right)-2cos^2a\)
\(=2-sin^2a-cos^2a=2-1=1\)
hình như đề sai hay sao ấy
tách mãi mà vẫn cứ phụ thuộc
đặt \(\sin\left(a\right)^2=x;\cos\left(a\right)^2=y;x+y=1\)
Ta có:
\(N=\sqrt{x^2+4y+\sqrt{y^2+4x}}=\sqrt{x^2+4\left(1-x\right)+\sqrt{y^2-4\left(1-y\right)}}\)
\(=\sqrt{x^2-4x+4+\sqrt{y^2-4y+4}}=\sqrt{\left(x-2\right)^2+\sqrt{\left(y-2\right)^2}}=\sqrt{\left(x-2\right)^2+\sqrt{\left(1-x-2\right)^2}}=\sqrt{\left(x-2\right)^2+\sqrt{\left(x+1\right)^2}}\)\(=\sqrt{x^2-4x+4+x+1}=\sqrt{x^2-3x+5}\)
Lời giải:
\(M=\frac{\frac{\sin a}{\cos a}+1}{\frac{\sin a}{\cos a}-1}=\frac{\tan a+1}{\tan a-1}=\frac{\frac{3}{5}+1}{\frac{3}{5}-1}=-4\)
\(N = \frac{\frac{\sin a\cos a}{\cos ^2a}}{\frac{\sin ^2a-\cos ^2a}{\cos ^2a}}=\frac{\frac{\sin a}{\cos a}}{(\frac{\sin a}{\cos a})^2-1}=\frac{\tan a}{\tan ^2a-1}=\frac{\frac{3}{5}}{\frac{3^2}{5^2}-1}=\frac{-15}{16}\)
a) Cần chứng minh \(\dfrac{1-cos\alpha}{sin\alpha}=\dfrac{sin\alpha}{1+cos\alpha}\)
\(\Rightarrow sin^2\alpha=\left(1-cos\alpha\right)\left(1+cos\alpha\right)\Rightarrow sin^2\alpha=1-cos^2\alpha\)
\(\Rightarrow sin^2\alpha+cos^2\alpha=1\)
Giả sử tam giác ABC vuông tại A
Ta có: \(\left\{{}\begin{matrix}sin^2B=\dfrac{AC^2}{BC^2}\\cos^2B=\dfrac{AB^2}{BC^2}\end{matrix}\right.\Rightarrow sin^2B+cos^2B=\dfrac{AC^2+AB^2}{BC^2}=\dfrac{BC^2}{BC^2}=1\)
a)\(\dfrac{1-cosa}{sina}=\dfrac{sina}{1+cosa}\)
<=>\(\left(1-cosa\right)\left(1+cosa\right)=sin^2a\)
<=>\(1-cos^2a=sin^2a\) (lđ)
b)Ta có VT=\(\dfrac{cosa}{1+sina}+tga=\dfrac{cosa}{1+sina}+\dfrac{sina}{cosa}=\dfrac{cos^2a+sin^2a+sina}{\left(1+sina\right)cosa}=\dfrac{1+sina}{\left(1+sina\right)cosa}=\dfrac{1}{cosa}=vp\left(dpcm\right)\)
Ta có: \(cot\alpha=\dfrac{cos\alpha}{sin\alpha}=\dfrac{cos^2\alpha}{sin\alpha.cos\alpha}=\sqrt{5}\)
Lại có: \(\dfrac{1}{cot\alpha}=tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{sin^2\alpha}{cos\alpha.sin\alpha}=\dfrac{1}{\sqrt{5}}\)
\(\Rightarrow A=\dfrac{cos^2\alpha}{sin\alpha.cos\alpha}+\dfrac{sin^2\alpha}{sin\alpha.cos\alpha}=\sqrt{5}+\dfrac{1}{\sqrt{5}}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)
Ta có : cot α = \(\sqrt{5}\Rightarrow\dfrac{cos\alpha}{sin\alpha}=\sqrt{5}\Rightarrow cos\alpha=\sqrt{5}.sin\alpha\)
\(A=\dfrac{sin^2\alpha+cos^2\alpha}{sin\alpha.cos\alpha}\)
\(A=\dfrac{sin^2\alpha+\left(\sqrt{5}sin\alpha\right)^2}{sin\alpha.\sqrt{5}sin\alpha}=\dfrac{sin^2\alpha+5sin^2\alpha}{\sqrt{5}sin^2\alpha}\)
\(A=\dfrac{6sin^2\alpha}{\sqrt{5}sin^2\alpha}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\)
\(\left(\sqrt{\dfrac{1+sin\alpha}{1-sin\alpha}}+\sqrt{\dfrac{1-sin\alpha}{1+sin\alpha}}\right).\dfrac{1}{\sqrt{1+tan^2\alpha}}\)
\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{\left(1-sin\alpha\right)\left(1+sin\alpha\right)}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{\left(1+sin\alpha\right)\left(1-sin\alpha\right)}}\right).\dfrac{1}{\sqrt{1+\left(\dfrac{sin\alpha}{cos\alpha}\right)^2}}\)
\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{1-sin^2\alpha}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{1-sin^2\alpha}}\right).\dfrac{1}{\sqrt{\dfrac{cos^2\alpha+sin^2\alpha}{cos^2\alpha}}}\)
\(=\left(\sqrt{\dfrac{\left(1+sin\alpha\right)^2}{cos^2\alpha}}+\sqrt{\dfrac{\left(1-sin\alpha\right)^2}{cos^2\alpha}}\right).\dfrac{1}{\sqrt{\dfrac{1}{cos^2\alpha}}}\)
\(=\left(\dfrac{1+sin\alpha}{cos\alpha}+\dfrac{1-sin\alpha}{cos\alpha}\right).\dfrac{1}{\dfrac{1}{cos\alpha}}=\dfrac{2}{cos\alpha}.cos\alpha=2\)