Đáp án: $B=1$
$C=2$
Giải thích các bước giải:
Ta có:
$B=\sin^4\alpha+\sin^2\alpha\cos^2\alpha+\cos^2\alpha$
$\to B=\sin^2\alpha(\sin^2\alpha+\cos^2\alpha)+\cos^2\alpha$
$\to B=\sin^2\alpha\cdot 1+\cos^2\alpha$
$\to B=\sin^2\alpha+\cos^2\alpha$
$\to B=1$
Ta có:
$C=\dfrac{1}{1+\sin\alpha}+\dfrac1{1-\sin\alpha}-2\tan^2\alpha$
$\to C=\dfrac{1-\sin\alpha+1+\sin\alpha}{(1-\sin\alpha)(1+\sin\alpha)}-2\tan^2\alpha$
$\to C=\dfrac{2}{1-\sin^2\alpha}-2\tan^2\alpha$
$\to C=\dfrac{2(\sin^2\alpha+\cos^2\alpha)}{\cos^2\alpha}-2\tan^2\alpha$
$\to C=\dfrac{2\sin^2\alpha+2\cos^2\alpha)}{\cos^2\alpha}-2\tan^2\alpha$
$\to C=2\cdot \dfrac{\sin^2\alpha}{\cos^2\alpha}+2-2\tan^2\alpha$
$\to C=2\cdot\tan^2\alpha+2-2\tan^2\alpha$
$\to C=2$
