\(\left(1+\frac{\sin^2}{\cos^2}\right)cos^2-\left(1+\frac{cos^2}{sin^2}\right)sin^2.\)

=> \(\frac{cos^2+sin^2}{cos^2}\left(cos^2\right)-\frac{sin^2+cos^2}{sin^2}\left(sin^2\right)\)

=> 1-1 =0

\(=\frac{1}{cos^2a}\cdot cos^2a+\frac{1}{sin^2a}\cdot sin^2a\) 

\(=1+1\) 

\(=2\)

\(a=\left(\frac{sina+\frac{sina}{cosa}}{cosa+1}\right)^2+1=\left(\frac{sina\left(cosa+1\right)}{cosa\left(cosa+1\right)}\right)^2+1\)

\(=tan^2a+1=\frac{1}{cos^2a}\)

\(b=\frac{sina}{cosa}\left(\frac{1+cos^2a-sin^2a}{sina}\right)=\frac{sina}{cosa}\left(\frac{2cos^2a}{sina}\right)=2cosa\)

\(c=1-\frac{cos^2a}{cot^2a}+\frac{sina.cosa}{\frac{cosa}{sina}}=1-cos^2a.\frac{sin^2a}{cos^2a}+\frac{sin^2a.cosa}{cosa}\)

\(=1-sin^2a+sin^2a=1\)

phần chứng minh biểu thức không phụ thuộc \(x\)

ta có : \(A=\dfrac{cot^2a-cos^2a}{cot^2a}+\dfrac{sinacosa}{cota}=\dfrac{cot^2a-cos^2a}{cot^2a}+\dfrac{cos^2a}{cot^2a}\)

\(=\dfrac{cot^2a-cos^2a+cos^2a}{cot^2a}=\dfrac{cot^2a}{cot^2a}=1\left(đpcm\right)\)

ý còn lại : xem lại đề nha bn

phần chứng minh đẳng thức

ta có : \(\dfrac{sin2a-2sina}{sin2a+2sina}+tan^2\dfrac{a}{2}=\dfrac{2sinacosa-2sina}{2sinacosa+2sina}+tan^2\dfrac{a}{2}\)

\(=\dfrac{2sina\left(cosa-1\right)}{2sina\left(cosa+1\right)}+tan^2\dfrac{a}{2}=\dfrac{cosa-1}{cosa+1}+tan^2\dfrac{a}{2}\)

\(=\dfrac{1-2sin^2\dfrac{a}{2}-1}{2cos^2\dfrac{a}{2}-1+1}+tan^2\dfrac{a}{2}=\dfrac{-2sin^2\dfrac{a}{2}}{2cos^2\dfrac{a}{2}}+tan^2\dfrac{a}{2}\)

\(=-tan^2\dfrac{a}{2}+tan^2\dfrac{a}{2}=0\left(đpcm\right)\)

ta có : \(\dfrac{sina}{1+cosa}+\dfrac{1+cosa}{sina}=\dfrac{sin^2a+\left(1+cosa\right)^2}{sina\left(1+cosa\right)}\)

\(=\dfrac{sin^2a+cos^2a+2cosa+1}{sina\left(1+cosa\right)}=\dfrac{2cosa+2}{sina\left(cosa+1\right)}\)

\(=\dfrac{2\left(cosa+1\right)}{sina\left(cosa+1\right)}=\dfrac{2}{sina}\left(đpcm\right)\)

còn 2 câu kia để chừng nào rảnh mk giải cho nha

mk lm 2 câu còn lại nha

ta có : \(\dfrac{sin^2x}{sinx-cosx}-\dfrac{sinx+cosx}{tan^2x-1}=\dfrac{\left(1-cos^2x\right)\left(tan^2x-1\right)-\left(sin^2x-cos^2x\right)}{\left(sinx-cosx\right)\left(tan^2x-1\right)}\)

\(=\dfrac{tan^2x-sin^2x-sin^2x-sin^2x+cos^2x}{\left(sinx-cosx\right)\left(tan^2x-1\right)}=\dfrac{\dfrac{sin^4x}{cos^2x}-sin^2x-sin^2x+cos^2x}{\left(sinx-cosx\right)\left(tan^2-1\right)}\)

\(=\dfrac{tan^2x\left(sin^2x-cos^2x\right)-\left(sin^2x-cos^2x\right)}{\left(sinx-cosx\right)\left(tan^2x-1\right)}=\dfrac{\left(tan^2x-1\right)\left(sin^2x-cos^2x\right)}{\left(sinx-cosx\right)\left(tan^2x-1\right)}\)

\(=sinx+cosx\left(đpcm\right)\)

ta có : \(\dfrac{sin\left(a+b\right)sin\left(a-b\right)}{1-tan^2a.cot^2b}=\dfrac{sin\left(a+b\right)sin\left(a-b\right)}{1-\dfrac{sin^2a.cos^2b}{cos^2a.sin^2b}}\)

\(=\dfrac{sin\left(a+b\right)sin\left(a-b\right)}{\dfrac{cos^2a.sin^2b-sin^2a.cos^2b}{cos^2a.sin^2b}}=\dfrac{sin\left(a+b\right)sin\left(a-b\right).cos^2a.sin^2b}{-\left(sin^2a.cos^2b-cos^2a.sin^2b\right)}\)

\(=\dfrac{sin\left(a+b\right)sin\left(a-b\right).cos^2a.sin^2b}{-\left(\left(sina.cosb-cosa.sinb\right)\left(sina.cosb+cosa.sinb\right)\right)}\)

\(=\dfrac{sin\left(a+b\right)sin\left(a-b\right).cos^2a.sin^2b}{-sin\left(a-b\right)sin\left(a+b\right)}=-cos^2a.sin^2b\left(đpcm\right)\)

mk lm hơi tắc ! do tối rồi , mà mk lại đang ở quán nek nên không tiện làm dài . bạn thông cảm

Lời giải:

a)

\(\frac{\sin ^2a+2\cos ^2a-1}{\cot ^2a}=\frac{(\sin ^2a+\cos ^2a)+\cos ^2a-1}{\cot ^2a}=\frac{1+\cos ^2a-1}{\cot ^2a}=\frac{\cos ^2a}{\cot ^2a}=\frac{\cos ^2a}{(\frac{\cos a}{\sin a})^2}=\sin ^2a\)

b)

\(\frac{1-\sin ^2a\cos ^2a}{\cos ^2a}-\cos ^2a=\frac{1}{\cos ^2a}-\sin ^2a-\cos ^2a\)

\(=\frac{\sin ^2a+\cos ^2a}{\cos ^2a}-(\sin ^2a+\cos ^2a)=\tan ^2a+1-1=\tan ^2a\)

c)

\(\frac{\sin ^2a-\tan ^2a}{\cos ^2a-\cot ^2a}=\frac{\sin ^2a-\frac{\sin ^2a}{\cos ^2a}}{\cos ^2a-\frac{\cos ^2a}{\sin ^2a}}=\frac{\sin ^4a(\cos ^2a-1)}{\cos ^4a(\sin ^2a-1)}\)

\(=\frac{\sin ^4a(-\sin ^2a)}{\cos ^4a(-\cos ^2a)}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)

Câu a)

Từ \(\tan a=3\Leftrightarrow \frac{\sin a}{\cos a}=3\Rightarrow \sin a=3\cos a\)

Do đó:

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

\(=\frac{\cos ^2a(3+1)}{\cos ^2a(18-1)}=\frac{4}{17}\)

Câu b)

Có: \(\cot \left(\frac{\pi}{2}-x\right)=\tan x=\frac{\sin x}{\cos x}\)

\(\cos\left(\frac{\pi}{2}+x\right)=-\sin x\)

\(\Rightarrow \cot \left(\frac{\pi}{2}-x\right)\cos \left(\frac{\pi}{2}+x\right)=\frac{-\sin ^2x}{\cos x}\)

Và:

\(\frac{\sin (\pi-x)\cot x}{1-\sin ^2x}=\frac{\sin x\cot x}{\cos^2x}=\frac{\sin x.\frac{\cos x}{\sin x}}{\cos^2x}=\frac{1}{\cos x}\)

Do đó:

\(\Rightarrow \cot \left(\frac{\pi}{2}-x\right)\cos \left(\frac{\pi}{2}+x\right)+\frac{\sin (\pi-x)\cot x}{1-\sin ^2x}=\frac{1-\sin ^2x}{\cos x}=\frac{\cos ^2x}{\cos x}=\cos x\)

Ta có đpcm.

~ ~ ~ Áp dụng đẳng thức \(\left(a+b\right)^2+\left(a-b\right)^2=2\left(a^2+b^2\right)\) ~ ~ ~

a)

\(\left(\sin\alpha+\cos\alpha\right)^2-2\sin\alpha\cos\alpha-1\)

\(=\left(\sin\alpha+\cos\alpha\right)^2-\left(2\sin\alpha\cos\alpha+\sin^2\alpha+\cos^2\alpha\right)\)

\(=\left(\sin\alpha+\cos\alpha\right)^2-\left(\sin\alpha+\cos\alpha\right)^2\)

= 0

b)

\(\left(\sin\alpha-\cos\alpha\right)^2+2\sin\alpha\cos\alpha+1\)

\(=\left(\sin\alpha-\cos\alpha\right)^2+2\sin\alpha\cos\alpha+\sin^2\alpha+\cos^2\alpha\)

\(=\left(\sin\alpha-\cos\alpha\right)^2+\left(\sin\alpha+\cos\alpha\right)^2\)

\(=2\left(\sin^2\alpha+\cos^2\alpha\right)\)

= 2

c)

\(\left(\sin\alpha+\cos\alpha\right)^2+\left(\sin\alpha-\cos\alpha\right)^2+2\)

\(=2\left(\sin^2\alpha+\cos^2\alpha\right)+2\)

= 4

d)

\(\sin^2\alpha\cot^2\alpha+\cos^2\alpha\tan^2\alpha\)

\(=\left(\sin\times\dfrac{\cos}{\sin}\right)^2+\left(\cos\times\dfrac{\sin}{\cos}\right)^2\)

= 1

Ta có:

 \(W_t=\dfrac{1}{2}m\omega^2A^2cos^2\left(\omega t+\varphi_0\right)\\ W_d=\dfrac{1}{2}mv^2=\dfrac{1}{2}m\omega^2A^2sin^2\left(\omega t+\varphi_0\right)\\ \Rightarrow W=W_t+W_d=\dfrac{1}{2}m\omega^2A^2\left[cos^2\left(\omega t+\varphi_0\right)+sin^2\left(\omega t+\varphi_0\right)\right]\\ \Rightarrow W=\dfrac{1}{2}m\omega^2A^2\)