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D = \(\left(sin^2a+cos^2a\right)+\left(cos\left(90-a\right)-sina\right)+1+\left(tan^2\left(90-a\right)-\frac{1}{sin^2a}\right)\)
\(=1+\left(sina-sina\right)+1+\left(cot^2a-1-cos^2a\right)=1+1-1=1\)
A = \(\left(sin^2a+cos^2a\right)^2=1^2=1\)
D = \(sin^2\left(sin^2B+cos^2B\right)+cos^2a=sin^2a+cos^2a=1\)
a/ \(A=\left(sin\alpha+cos\alpha\right)^2+\left(sin\alpha-cos\alpha\right)^2=2\left(sin^2\alpha+cos^2\alpha\right)=2\)
b/ \(B=\left(1+tan^2\alpha\right)\left(1-sin^2\alpha\right)-\left(1+cotg^2\alpha\right)\left(1-cos^2\alpha\right)\)
\(=\left(1+\frac{sin^2\alpha}{cos^2\alpha}\right)\left(1-sin^2\alpha\right)-\left(1+\frac{cos^2\alpha}{sin^2\alpha}\right)\left(1-cos^2\alpha\right)\)
\(=\frac{1}{cos^2\alpha}.cos^2\alpha-\frac{1}{sin^2\alpha}.sin^2\alpha=1-1=0\)
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: \(1+tan^2a=\dfrac{1}{cos^2a}=1+\dfrac{9}{25}=\dfrac{34}{25}\)
=>cos^2a=25/34
=>\(cosa=\dfrac{5}{\sqrt{34}}\)
\(sina=\sqrt{1-\dfrac{25}{34}}=\dfrac{3}{\sqrt{34}}\)
\(A=\left(\sin\alpha+\cos\alpha+\sin\alpha-\cos\alpha\right)^2-2\left(\sin\alpha+\cos\alpha\right)\left(\sin\alpha-\cos\alpha\right)\)
\(=4\sin^2\alpha-2\sin^2\alpha+2\cos^2\alpha=2\left(\sin^2\alpha+\cos^2\alpha\right)=2\)
\(B=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)=\sin^4\alpha+\cos^4\alpha+2\sin^2\alpha.\cos^2\alpha\)
\(=\left(\sin^2\alpha+\cos^2\alpha\right)^2-1=0\)
\(C=3\left(\sin^4\alpha+\cos^4\alpha\right)-2\sin^2\alpha.\cos^2\alpha\left(\sin^2\alpha+\cos^2\alpha\right)=3\left(\sin^4\alpha+\cos^4\alpha\right)-2\sin^2\alpha.\cos^2\alpha\)
\(=3\left(\sin^2\alpha+\cos^2\alpha-\frac{1}{9}\right)^2-\frac{1}{9}=\frac{61}{27}\)
Câu1. Ta có\(\sin^2\alpha+\cos^2\alpha=1\Leftrightarrow\sin^2\alpha=1-\cos^2\alpha=1-\left(\frac{1}{4}\right)^2\)
\(=\frac{15}{16}\Rightarrow\sin\alpha=\frac{\sqrt{15}}{4}\)
\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\sqrt{15}}{4}:\frac{1}{4}=\sqrt{15}\)\(=4\sin\alpha\)
Câu2.
Ta có: \(\frac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}=5\Leftrightarrow\cos\alpha+\sin\alpha=5\cos\alpha-5\sin\alpha\)
\(\Leftrightarrow4\cos\alpha=6\sin\alpha\Leftrightarrow\frac{\sin\alpha}{\cos\alpha}=\frac{2}{3}\)
\(\Rightarrow\tan\alpha=\frac{2}{3}\)
Bài 1 :
\(C=cos^2a\left(cos^2a+sin^2a\right)+sin^2a=cos^2a+sin^2a=1\)
b: Xét ΔADC vuông tại D và ΔBEC vuông tại E có
\(\widehat{C}\) chung
Do đó: ΔADC\(\sim\)ΔBEC