Giúp mình với CMR a, tg^2a + 1 = 1/ cos^2a b, cotg^2a + 1 = 1/sin^2a 12/08/2021 Bởi Allison Giúp mình với CMR a, tg^2a + 1 = 1/ cos^2a b, cotg^2a + 1 = 1/sin^2a
a, $\tan^2a+1$ $=\dfrac{\sin^2a}{\cos^2a}+1$ $=\dfrac{\sin^2a+\cos^2a}{\cos^2a}$ $=\dfrac{1}{\cos^2a}$ b, $\cot^2a+1$ $=\dfrac{\cos^2a}{\sin^2a}+1$ $=\dfrac{\cos^2a+\sin^2a}{\sin^2a}$ $=\dfrac{1}{\sin^2a}$ Bình luận
Đáp án: Giải thích các bước giải: $a,tg^2\alpha+1=\dfrac{sin^2\alpha}{cos^2\alpha}+\dfrac{cos^2\alpha}{cos^2\alpha}=\dfrac{sin^2\alpha+cos^2\alpha}{cos^2\alpha}=\dfrac{1}{cos^2\alpha}(đpcm)$ $b,cotg^2\alpha+1=\dfrac{cos^2\alpha}{sin^2\alpha}+\dfrac{sin^2\alpha}{sin^2\alpha}=\dfrac{sin^2\alpha+cos^2\alpha}{sin^2\alpha}=\dfrac{1}{sin^2\alpha}(đpcm)$ Bình luận
a,
$\tan^2a+1$
$=\dfrac{\sin^2a}{\cos^2a}+1$
$=\dfrac{\sin^2a+\cos^2a}{\cos^2a}$
$=\dfrac{1}{\cos^2a}$
b,
$\cot^2a+1$
$=\dfrac{\cos^2a}{\sin^2a}+1$
$=\dfrac{\cos^2a+\sin^2a}{\sin^2a}$
$=\dfrac{1}{\sin^2a}$
Đáp án:
Giải thích các bước giải:
$a,tg^2\alpha+1=\dfrac{sin^2\alpha}{cos^2\alpha}+\dfrac{cos^2\alpha}{cos^2\alpha}=\dfrac{sin^2\alpha+cos^2\alpha}{cos^2\alpha}=\dfrac{1}{cos^2\alpha}(đpcm)$
$b,cotg^2\alpha+1=\dfrac{cos^2\alpha}{sin^2\alpha}+\dfrac{sin^2\alpha}{sin^2\alpha}=\dfrac{sin^2\alpha+cos^2\alpha}{sin^2\alpha}=\dfrac{1}{sin^2\alpha}(đpcm)$