Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{\cos\alpha}{1-\sin\alpha}=\frac{1+\sin\alpha}{\cos\alpha}\Leftrightarrow\cos^2\alpha=1-\sin^2\alpha\)\(\Leftrightarrow\cos^2\alpha+\sin^2\alpha=1\)(luôn đúng)
\(\frac{\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha-\cos\alpha\right)^2}{\sin\alpha\cdot\cos\alpha}=\frac{\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cdot\cos\alpha-\sin^2\alpha-\cos^2\alpha+2\sin\alpha\cdot\cos\alpha}{\sin\alpha\cdot\cos\alpha}\)
\(=\frac{4\sin\alpha\cdot\cos\alpha}{\sin\alpha\cdot\cos\alpha}=4\)(đpcm)
a: \(\sin^2a+\cos^2a=1\)
\(\Leftrightarrow\cos^2a=1-\sin^2a=\left(1-\sin a\right)\left(1+\sin a\right)\)
hay \(\dfrac{\cos a}{1-\sin a}=\dfrac{1+\sin a}{\cos a}\)
b: \(VT=\dfrac{\left(\sin a+\cos a+\sin a-\cos a\right)\left(\sin a+\cos a-\sin a+\cos a\right)}{\sin a\cdot\cos a}\)
\(=\dfrac{2\cdot\cos a\cdot2\sin a}{\sin a\cdot\cos a}=4\)
Câu 1:
\(\cos a=\sqrt{1-\left(\dfrac{1}{4}\right)^2}=\dfrac{\sqrt{15}}{4}\)
\(A=\sin^2a+3\cos^2a-1=\dfrac{1}{16}+3\cdot\dfrac{15}{16}-1=\dfrac{15}{8}\)
VT=\(\dfrac{c\text{os}a}{1-sina}\)
\(=\dfrac{c\text{os}a\left(1+sina\right)}{\left(1-sina\right)\left(1+sina\right)}=\dfrac{c\text{os}a\left(1+sina\right)}{1-sin^2a}\\ \\ \\ =\dfrac{c\text{os}a\left(1+sina\right)}{c\text{os}^2a}=\dfrac{1+sina}{c\text{os}a}=VP\left(\text{đ}pcm\right)\)
\(\dfrac{\left(cosa-sina\right)^2-\left(cosa+sina\right)^2}{cosa\cdot sina}\)
\(=\dfrac{\left(cosa-sina-cosa-sina\right)\left(cosa-sina+cosa+sina\right)}{cosa\cdot sina}\)
\(=\dfrac{-2\cdot sina\cdot2\cdot cosa}{cosa\cdot sina}=-4\)
Lời giải:
Ta có:
\(A=\tan ^3a+\cot ^3a=\frac{\sin ^3a}{\cos ^3a}+\frac{\cos ^3a}{\sin ^3a}\)
\(=\frac{(\sin a)^6+(\cos a)^6}{(\sin a\cos a)^3}\)
\(=\frac{(\sin ^2a+\cos ^2a)(\sin ^4a-\sin ^2a\cos ^2a+\cos ^4a)}{(\sin a\cos a)^3}\)
\(=\frac{\sin^4 a-\sin ^2a\cos ^2a+\cos ^4a}{(\sin a\cos a)^3}\)
\(=\frac{(\sin ^2a+\cos ^2a)^2-3\sin ^2a\cos ^2a}{(\sin a\cos a)^3}=\frac{1-3(\sin a\cos a)^2}{(\sin a\cos a)^3}(*)\)
Mặt khác: \(\sin a+\cos a=1,366\)
\(\Rightarrow \sin ^2a+2\sin a\cos a+\cos ^2a=1,366^2\)
\(\Rightarrow 2\sin a\cos a=1,366^2-1\Rightarrow \sin a\cos a=\frac{1,366^2-1}{2}\)
Thay vào A ở $(*)$ suy ra:
\(A\approx 5,391\)
\(=\frac{\left(\sin a+\cos a-\sin a+\cos a\right)\left(\sin a+\cos a+\sin a-\cos a\right)}{\sin a.\cos a}=\frac{2.\cos a.2.\sin a}{\sin a.\cos a}=4\)