a) $sin^2a.tan^2a+4sin^2a-tan^2a+3cos^2a=3$;
b) $cotx-tanx-2tan2x-4tan4x=8cot8x$;
c) $sin^2x+tan^2x=\frac{1}{cos^2x}-cos^2x$
c) $\sin^{2}x+\tan^{2}x=\sin^{2}x+\dfrac{\sin^{2}x}{\cos^{2}x}=\dfrac{\sin^{2}x\cos^{2}x+\sin^{2}x}{\cos^{2}x}$

$=\dfrac{\sin^{2}x(1+\cos^{2}x)}{\cos^{2}x}=\dfrac{(1-\cos^{2}x)(1+\cos^{2}x)}{\cos^{2}x}$

$=\dfrac{1-\cos^{4}x}{\cos^{2}x}=\dfrac{1}{\cos^{2}x}-\cos^{2}x$
b) $\cot x-\tan x=\dfrac{\cos x}{\sin x}-\dfrac{\sin x}{\cos x}=\dfrac{\cos^{2}x-\sin^{2}x}{\sin x\cos x}$

$=\dfrac{\cos 2x}{\dfrac{\sin 2x}{2}}=2\cot 2x$

Áp dụng ta có:

$\cot x-\tan x-2\tan 2x-4\tan 4x=(\cot x-\tan x)-2\tan 2x-4\tan 4x$

$=2\cot 2x-2\tan 2x-4\tan 4x=2(\cot 2x-\tan 2x)-4\tan 4x$

$=4\cot 4x-4\tan 4x=4(\cot 4x-\tan 4x)=8\cot 8x$
a) $\sin^{2}a\tan^{2}a+4\sin^{2}a-\tan^{2}a+3\cos ^{2}a$

$=(\sin^{2}a\tan^{2}a+\sin^{2}a)-\tan^{2}a+3(\sin^{2}a+\cos^{2}a)$

$=\sin^{2}a(\tan^{2}a+1)-\tan^{2}a+3$

$=\sin^{2}a.\dfrac{1}{\cos^{2}a}-\tan^{2}a+3=\tan^{2}a-\tan^{2}a+3=3$

Bạn cần đăng nhập để có thể gửi đáp án

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