Question

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Answer 1

a, $c o t x + \tan x = \frac{2}{\sin 2 x}$

VT: $c o t x + \tan x = \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} = \frac{\cos^{2} x + \sin^{2} x}{\sin x . \cos x} = \frac{1}{\frac{1}{2} . \sin 2 x} = \frac{2}{\sin 2 x}$

$\Rightarrow V T = V P (d p c m)$

b, $c o t x - \tan x = 2 c o t 2 x$

VT: $c o t x - \tan x = \frac{c o s x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^{2} x - \sin^{2} x}{s i n x . \cos x} = \frac{\cos 2 x}{\frac{1}{2} . \sin 2 x} = 2 c o t 2 x$

$\Rightarrow V T = V P (đ p c m)$

c, $\frac{\sin 2 x}{1 + \cos 2 x} = \tan x$

VT: $\frac{2 \sin x \cos x}{\sin^{2} x + \cos^{2} x + \cos^{2} x - \sin^{2} x} = \frac{2 \sin x \cos x}{2 \cos^{2} x} = \frac{\sin x}{\cos x} = \tan x$

$\Rightarrow V T = V P (đ p c m)$

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Answer 2

a, $c o t x + \tan x = \frac{2}{\sin 2 x}$

VT: $c o t x + \tan x = \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} = \frac{\cos^{2} x + \sin^{2} x}{\sin x . \cos x} = \frac{1}{\frac{1}{2} . \sin 2 x} = \frac{2}{\sin 2 x}$

$\Rightarrow V T = V P (d p c m)$

b, $c o t x - \tan x = 2 c o t 2 x$

VT: $c o t x - \tan x = \frac{c o s x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^{2} x - \sin^{2} x}{s i n x . \cos x} = \frac{\cos 2 x}{\frac{1}{2} . \sin 2 x} = 2 c o t 2 x$

$\Rightarrow V T = V P (đ p c m)$

c, $\frac{\sin 2 x}{1 + \cos 2 x} = \tan x$

VT: $\frac{2 \sin x \cos x}{\sin^{2} x + \cos^{2} x + \cos^{2} x - \sin^{2} x} = \frac{2 \sin x \cos x}{2 \cos^{2} x} = \frac{\sin x}{\cos x} = \tan x$

$\Rightarrow V T = V P (đ p c m)$

Answer 3

d,

$V T = \frac{2 \cos^{2} x \times \cos^{2} 2 x}{\cos^{2} 2 x \times 2 \sin x \times \cos x} = \frac{\cos x}{\sin x} = c o t x (đ p c m)$

Answer 4

a, $c o t x + \tan x = \frac{2}{\sin 2 x}$

VT: $c o t x + \tan x = \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} = \frac{\cos^{2} x + \sin^{2} x}{\sin x . \cos x} = \frac{1}{\frac{1}{2} . \sin 2 x} = \frac{2}{\sin 2 x}$

$\Rightarrow V T = V P (d p c m)$

b, $c o t x - \tan x = 2 c o t 2 x$

VT: $c o t x - \tan x = \frac{c o s x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^{2} x - \sin^{2} x}{s i n x . \cos x} = \frac{\cos 2 x}{\frac{1}{2} . \sin 2 x} = 2 c o t 2 x$

$\Rightarrow V T = V P (đ p c m)$

c, $\frac{\sin 2 x}{1 + \cos 2 x} = \tan x$

VT: $\frac{2 \sin x \cos x}{\sin^{2} x + \cos^{2} x + \cos^{2} x - \sin^{2} x} = \frac{2 \sin x \cos x}{2 \cos^{2} x} = \frac{\sin x}{\cos x} = \tan x$

$\Rightarrow V T = V P (đ p c m)$

...Xem thêm

Answer 5

a, $c o t x + \tan x = \frac{2}{\sin 2 x}$

VT: $c o t x + \tan x = \frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} = \frac{\cos^{2} x + \sin^{2} x}{\sin x . \cos x} = \frac{1}{\frac{1}{2} . \sin 2 x} = \frac{2}{\sin 2 x}$

$\Rightarrow V T = V P (d p c m)$

b, $c o t x - \tan x = 2 c o t 2 x$

VT: $c o t x - \tan x = \frac{c o s x}{\sin x} - \frac{\sin x}{\cos x} = \frac{\cos^{2} x - \sin^{2} x}{s i n x . \cos x} = \frac{\cos 2 x}{\frac{1}{2} . \sin 2 x} = 2 c o t 2 x$

$\Rightarrow V T = V P (đ p c m)$

c, $\frac{\sin 2 x}{1 + \cos 2 x} = \tan x$

VT: $\frac{2 \sin x \cos x}{\sin^{2} x + \cos^{2} x + \cos^{2} x - \sin^{2} x} = \frac{2 \sin x \cos x}{2 \cos^{2} x} = \frac{\sin x}{\cos x} = \tan x$

$\Rightarrow V T = V P (đ p c m)$

3 câu trả lời 1815

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