Toán Sin^a+sin^a.tan^a Các cao nhâb giúp em vs ạ 14/07/2021 By Eva Sin^a+sin^a.tan^a Các cao nhâb giúp em vs ạ
Đáp án: $\sin^2a+\sin^2a.\tan^2a=\tan^2a$ Giải thích các bước giải: $\sin^2a+\sin^2a.\tan^2a$ $=\sin^2a+\sin^2a.\dfrac{\sin^2a}{\cos^2a}$ $=\sin^2a.\left(1+\dfrac{\sin^2a}{\cos^2a}\right)$ $=\sin^2a.\dfrac{\cos^2a+\sin^2a}{\cos^2a}$ $=\dfrac{\sin^2a}{\cos^2a}$ $=\tan^2a$. Trả lời
Đáp án:Sin^a+sin^a.tan^a=sin^a+sin^a.[sin^a/cos^a]=sin^a+sin^a.[sin^a/[1-sin^a^2]] =sin^a+ [sin^a^2/[1-sin^a^2]]=[sin^a-sin^a^2+sin^a^2]/[1-sin^a^2] =sin^a^2/[1-sin^a^2] Trả lời
Đáp án:
$\sin^2a+\sin^2a.\tan^2a=\tan^2a$
Giải thích các bước giải:
$\sin^2a+\sin^2a.\tan^2a$
$=\sin^2a+\sin^2a.\dfrac{\sin^2a}{\cos^2a}$
$=\sin^2a.\left(1+\dfrac{\sin^2a}{\cos^2a}\right)$
$=\sin^2a.\dfrac{\cos^2a+\sin^2a}{\cos^2a}$
$=\dfrac{\sin^2a}{\cos^2a}$
$=\tan^2a$.
Đáp án:Sin^a+sin^a.tan^a=sin^a+sin^a.[sin^a/cos^a]=sin^a+sin^a.[sin^a/[1-sin^a^2]]
=sin^a+ [sin^a^2/[1-sin^a^2]]=[sin^a-sin^a^2+sin^a^2]/[1-sin^a^2]
=sin^a^2/[1-sin^a^2]