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NV
29 tháng 10 2020

\(=\left(1+tan^220\right).cos^220-tan40.cot\left(90-50\right)\)

\(=\left(1+\frac{sin^220}{cos^220}\right).cos^220-tan40.cot40\)

\(=cos^220+sin^220-1\)

\(=1-1=0\)

21 tháng 11 2016

a/ \(\tan40.\cot40+\frac{\sin50}{\cos40}\)

\(=1+\frac{\cos40}{\cos40}=1+1=2\)

21 tháng 11 2016

Đề yêu cầu làm gì bạn

8 tháng 8 2018

\(A=sin^210+sin^220+sin^230+sin^280+sin^270+sin^260=sin^210+sin^220+sin^230+cos^210+cos^220+cos^230=1+1+1=3\)\(B=\left(1+tan^2\alpha\right)\left(1-sin^2\alpha\right)+\left(1+cot^2\alpha\right)\left(1-cos^2\alpha\right)=\dfrac{1}{cos^2\alpha}.cos^2\alpha+\dfrac{1}{sin^2\alpha}.sin^2\alpha=1+1=2\)

18 tháng 8 2018

ta có : \(5tan40.tan50-cos^247-3-cos^243\)

\(=5tan40.tan\left(90-40\right)-cos^247-cos^2\left(90-47\right)-3\)

\(=5.tan40.cot40-cos^247-sin^247-3=5-1-3=1\)

16 tháng 7 2019

ĐKXĐ:...

\(VT=\frac{\frac{\cos^2\left(\frac{x}{2}\right)}{\sin^2\left(\frac{x}{2}\right)}-\frac{\cos^2\left(\frac{3x}{2}\right)}{\sin^2\left(\frac{3x}{2}\right)}}{\cos^2\left(\frac{x}{2}\right).\cos x.\frac{1}{\sin^2\left(\frac{3x}{2}\right)}}\) \(=\frac{\sin^2\left(\frac{3x}{2}\right)}{\sin^2\left(\frac{x}{2}\right).\cos x}-\frac{\cos^2\left(\frac{3x}{2}\right)}{\cos^2\left(\frac{x}{2}\right).\cos x}\)

\(=\frac{\sin^2\left(\frac{3x}{2}\right).\cos^2\left(\frac{x}{2}\right)-\cos^2\left(\frac{3x}{2}\right).\sin^2\left(\frac{x}{2}\right)}{\sin^2\left(\frac{x}{2}\right).\cos x.\cos^2\left(\frac{x}{2}\right)}\) \(=\frac{\left(\sin\left(\frac{3x}{2}\right).\cos\left(\frac{x}{2}\right)-\cos\left(\frac{3x}{2}\right).\sin\left(\frac{x}{2}\right)\right).\left(\sin\left(\frac{3x}{2}\right).\cos\left(\frac{x}{2}\right)+\cos\left(\frac{3x}{2}\right).\sin\left(\frac{x}{2}\right)\right)}{\sin^2\left(\frac{x}{2}\right).\cos x.\cos^2\left(\frac{x}{2}\right)}\)

\(=\frac{\sin\left(\frac{3x}{2}-\frac{x}{2}\right).\sin\left(\frac{3x}{2}+\frac{x}{2}\right)}{\sin^2\left(\frac{x}{2}\right).\cos x.\cos^2\left(\frac{x}{2}\right)}=\frac{\sin x.\sin2x}{\sin^2\left(\frac{x}{2}\right).\cos x.\cos^2\left(\frac{x}{2}\right)}\)

\(=\frac{2.\sin^2x.\cos x}{\sin^2\left(\frac{x}{2}\right).\cos x.\cos^2\left(\frac{x}{2}\right)}=\frac{8.\sin^2\left(\frac{x}{2}\right).\cos^2\left(\frac{x}{2}\right)}{\sin^2\left(\frac{x}{2}\right).\cos^2\left(\frac{x}{2}\right)}=8\left(đpcm\right)\)

NV
1 tháng 10 2020

b.

\(\Leftrightarrow\frac{2\pi}{3}\left(sinx-1\right)=k2\pi\)

\(\Leftrightarrow sinx-1=3k\)

\(\Leftrightarrow sinx=3k+1\)

Do \(-1\le sinx\le1\)

\(\Rightarrow-1\le3k+1\le1\Rightarrow-\frac{2}{3}\le k\le0\)

\(\Rightarrow k=0\)

\(\Rightarrow sinx=1\)

\(\Rightarrow x=\frac{\pi}{2}+k2\pi\)

NV
1 tháng 10 2020

c.

ĐKXĐ: ...

\(\Leftrightarrow\frac{\pi}{4}\left(cosx-1\right)=-\frac{\pi}{4}+k\pi\)

\(\Leftrightarrow cosx-1=4k-1\)

\(\Leftrightarrow cosx=4k\)

\(-1\le cosx\le1\Rightarrow-1\le4k\le1\)

\(\Rightarrow-\frac{1}{4}\le k\le\frac{1}{4}\Rightarrow k=0\)

\(\Rightarrow cosx=0\)

\(\Rightarrow x=\frac{\pi}{2}+k\pi\)

NV
8 tháng 6 2020

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

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

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

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

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

NV
4 tháng 8 2020

\(A=\frac{sinx}{cosx}+\frac{cosx}{sinx}+\frac{sin3x}{cos3x}+\frac{cos3x}{sin3x}\)

\(=\frac{sin^2x+cos^2x}{sinx.cosx}+\frac{sin^23x+cos^23x}{sin3x.cos3x}=\frac{2}{2sinx.cosx}+\frac{2}{2sin3x.cos3x}\)

\(=\frac{2}{sin2x}+\frac{2}{sin6x}=\frac{2\left(sin2x+sin6x\right)}{sin2x.sin6x}=\frac{4sin4x.cos2x}{sin2x.sin6x}\)

\(=\frac{8sin2x.cos^22x}{sin2x.sin6x}=\frac{8cos^22x}{sin6x}\)

\(B=\frac{sin30}{cos30}+\frac{sin60}{cos60}+\frac{sin40}{cos40}+\frac{sin50}{cos50}=\frac{sin30.cos60+cos30.sin60}{cos30.cos60}+\frac{sin40.cos50+sin50.cos40}{cos40.cos50}\)

\(=\frac{sin90}{cos30.cos60}+\frac{sin90}{cos40.cos50}=\frac{1}{\frac{1}{2}.\frac{\sqrt{3}}{2}}+\frac{1}{\frac{1}{2}cos90+\frac{1}{2}cos10}\)

\(=\frac{4\sqrt{3}}{3}+\frac{2}{cos10}=\frac{4\sqrt{3}\left(cos10+\frac{\sqrt{3}}{2}\right)}{3cos10}=\frac{4\sqrt{3}\left(cos10+cos30\right)}{3cos10}\)

\(=\frac{8\sqrt{3}cos20.cos10}{3cos10}=\frac{8\sqrt{3}}{3}cos20\)

5 tháng 10 2018

Ta có : \(C=\left(1+\frac{11}{13}\right).\left(1+\frac{11}{28}\right).\left(1+\frac{11}{45}\right)......\left(1+\frac{11}{220}\right)\)

                \(=\frac{24}{13}.\frac{39}{28}.\frac{56}{45}......\frac{231}{220}\)

                \(=\frac{2.12}{1.13}.\frac{3.13}{2.14}.\frac{4.14}{3.15}......\frac{11.21}{10.22}\)

                \(=\frac{2.12.3.13.4.14.....11.21}{1.13.2.14.3.15.....10.22}\)

                \(=\frac{\left(2.3.4.....11\right).\left(12.13.14.....21\right)}{\left(1.2.3.....10\right).\left(13.14.15.....22\right)}\)

                \(=\frac{11.12}{1.22}=\frac{12}{1.2}=6\)

Vậy C = 6

6 tháng 10 2018

Nguyễn Thị Tiểu Ngân Lớp 7a thcs Dĩnh Kế phải ko ?