$1 + tan^2 b = \dfrac{1}{cos^2 b}$
$ \to tan^2 b = \dfrac{1}{cos^2 b}-1$
$\to tan^2 b = \dfrac{1- cos^2 b}{cos^2 b}$
$\to tan^2 b = \dfrac{sin^2 b}{cos^2 b}$
$\to tan^2 b = (\dfrac{sinb}{cosb})^2 = tan^2 b$ (đúng)
Vậy ta có điều phải chứng minh
$\displaystyle \begin{array}{{>{\displaystyle}l}} Ta\ có\ :\ tanB=\frac{sinB}{cosB} \ \\ \rightarrow \ tan^{2} B=\frac{sin^{2} B}{cos^{2} B} \ \\ \rightarrow 1+tan^{2} B=1+\frac{sin^{2} B}{cos^{2} B}\\ \rightarrow 1+tan^{2} B=\frac{cos^{2} B+sin^{2} B}{cos^{2} B} =\frac{1}{cos^{2} B} \end{array}$
Viết một đoạn văn trình bày cảm nhận của e về phong cách HCM Giúp e trc tối nay nha mn:3
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Giải giúp em với ạ câu 12 13 đến câu 25
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Giúp em đi . Hứa làm đủ . Bài yêu cầu tính ạ
Calli:" Draken-kun " 💖 =')))) Sợ hãi qua từng chap ☺💔 !!
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Giúp em đi màaaaaaaaaaaaaaaaaaaaaaaaaaaa
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