K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

11 tháng 5 2017

a) \(tan3\alpha-tan2\alpha-tan\alpha=\left(tan3\alpha-tan\alpha\right)-tan2\alpha\)
\(=\left(\dfrac{sin3\alpha}{cos3\alpha}-\dfrac{sin\alpha}{cos\alpha}\right)-\dfrac{sin2\alpha}{cos2\alpha}\)\(=\dfrac{sin3\alpha cos\alpha-cos3\alpha sin\alpha}{cos3\alpha cos\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\)
\(=\dfrac{sin2\alpha}{cos3\alpha cos\alpha}-\dfrac{sin2\alpha}{cos2\alpha}\)
\(=sin2\alpha.\left(\dfrac{1}{cos3\alpha cos\alpha}-\dfrac{1}{cos2\alpha}\right)\)
\(=sin2\alpha.\dfrac{cos2\alpha-cos3\alpha cos\alpha}{cos3\alpha cos\alpha cos2\alpha}\)
\(=sin2\alpha.\dfrac{cos2\alpha-\dfrac{1}{2}\left(cos4\alpha+cos2\alpha\right)}{cos3\alpha cos2\alpha cos\alpha}\)
\(=sin2\alpha.\dfrac{cos2\alpha-cos4\alpha}{2cos3\alpha cos2\alpha cos\alpha}\)
\(=\dfrac{sin2\alpha.2sin3\alpha.sin\alpha}{2cos3\alpha cos2\alpha cos\alpha}\)
\(=tan3\alpha tan2\alpha tan\alpha\) (Đpcm).

11 tháng 5 2017

b) \(\dfrac{4tan\alpha\left(1-tan^2\alpha\right)}{\left(1+tan^2\right)^2}=4tan\alpha\left(1-tan^2\alpha\right):\left(\dfrac{1}{cos^2\alpha}\right)^2\)
\(=4tan\alpha\left(1-tan^2\alpha\right)cos^4\alpha\)
\(=4\dfrac{sin\alpha}{cos\alpha}\left(1-\dfrac{sin^2\alpha}{cos^2\alpha}\right)cos^4\alpha\)
\(=4sin\alpha\left(cos^2\alpha-sin^2\alpha\right)cos\alpha\)
\(=4sin\alpha cos\alpha.cos2\alpha\)
\(=2.sin2\alpha.cos2\alpha=sin4\alpha\) (Đpcm).

NV
5 tháng 5 2020

\(tan\left(\frac{\pi}{3}-a\right)tan\left(\frac{\pi}{3}+a\right)=\frac{sin\left(\frac{\pi}{3}-a\right)sin\left(\frac{\pi}{3}+a\right)}{cos\left(\frac{\pi}{3}-a\right)cos\left(\frac{\pi}{3}+a\right)}\)

\(=\frac{cos2a-cos\frac{2\pi}{3}}{cos2a+cos\frac{2\pi}{3}}=\frac{cos2a+\frac{1}{2}}{cos2a-\frac{1}{2}}=\frac{2cos2a+1}{2cos2a-1}\)

\(\Rightarrow tana.tan\left(\frac{\pi}{3}-a\right)tan\left(\frac{\pi}{3}+a\right)=\frac{sina\left(2cos2a+1\right)}{cosa\left(2cos2a-1\right)}=\frac{2sina.cos2a+sina}{2cos2a.cosa-cosa}\)

\(=\frac{sin3a-sina+sina}{cos3a+cosa-cosa}=\frac{sin3a}{cos3a}=tan3a\)

2 tháng 3 2021

Điều kiện: a>45 độ

 

 

17 tháng 6 2016

a)\(tan3A=tan\left(A+2A\right)\)

\(=\frac{tanA+tan2A}{1-tanAtan2A}\)

\(=\frac{\frac{tanA+2tanA}{1-tan^2A}}{\frac{1-2tan^2A}{1-tan^2A}}\)

\(=\frac{\left(tanA-tan^3A+2tanA\right)}{1-tan^2A-2tan^2A}\)

\(=\frac{3tanA-tan^3A}{1-3tan^2A}\)

b)\(VT=cos^6A+sin^6A\)

\(=\left(cos^2A\right)^3+\left(sin^2A\right)^3\)

\(=\left(cos^2A+sin^2A\right)^3-3cos^2Asin^2A\left(cos^2A+sin^2A\right)^2\)

\(=1^3-3cos^2Asin^2A\left(1\right)^2\).Từ đó,\(sin^2A+cos^2A=1\)

\(=1-3cos^2Asin^2A=VP\)

18 tháng 6 2016

phần b tui sai

AH
Akai Haruma
Giáo viên
1 tháng 10 2018

a)

\(\sin ^4a-\cos ^4a+1=(\sin ^2a-\cos ^2a)(\sin ^2a+\cos^2a)+1\)

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

\(=2\sin ^2a\)

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

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

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

c)

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

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

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

AH
Akai Haruma
Giáo viên
1 tháng 10 2018

d)

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

\(=\frac{\sin ^2a.\frac{\cos ^2a-1}{\cos ^2a}}{\cos ^2a.\frac{\sin ^2a-1}{\sin ^2a}}\) \(=\frac{\sin ^2a.\frac{-\sin ^2a}{\cos ^2a}}{\cos ^2a.\frac{-\cos ^2a}{\sin ^2a}}=\frac{\sin ^6a}{\cos ^6a}=\tan ^6a\)

f)

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

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

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

NV
18 tháng 4 2021

\(VT=\dfrac{-tan\left(\dfrac{\pi}{2}-a\right)cos\left(2\pi-\dfrac{\pi}{2}+a\right)-sin^3\left(4\pi-\dfrac{\pi}{2}-a\right)}{cos\left(\dfrac{\pi}{2}-a\right)tan\left(2\pi-\dfrac{\pi}{2}+a\right)}\)

\(=\dfrac{-cota.sina+sin^3\left(\dfrac{\pi}{2}+a\right)}{sina.\left(-cota\right)}=\dfrac{-cosa+cos^3a}{-cosa}=1-cos^2a=sin^2a\)

10 tháng 5 2017

a) \(\dfrac{tan\alpha-tan\beta}{cot\beta-cot\alpha}=\dfrac{\dfrac{sin\alpha}{cos\alpha}-\dfrac{sin\beta}{cos\beta}}{\dfrac{cos\beta}{sin\beta}-\dfrac{cos\alpha}{sin\alpha}}\)
\(=\dfrac{\dfrac{sin\alpha cos\beta-cos\alpha sin\beta}{cos\alpha cos\beta}}{\dfrac{cos\beta sin\alpha-cos\alpha sin\beta}{sin\beta sin\alpha}}\)
\(=\dfrac{sin\beta sin\alpha}{cos\beta cos\alpha}=tan\alpha tan\beta\).

10 tháng 5 2017

b) \(tan100^o+\dfrac{sin530^o}{1+sin640^o}=tan100^o+\dfrac{sin170^o}{1+sin280^o}\)
\(=-cot10^o+\dfrac{sin10^o}{1-sin80^o}\)\(=\dfrac{-cos10^o}{sin10^o}+\dfrac{sin10^o}{1-cos10^o}\)
\(=\dfrac{-cos10^o+cos^210^o+sin^210^o}{sin10^o\left(1-cos10^o\right)}\) \(=\dfrac{1-cos10^o}{sin10^o\left(1-cos10^o\right)}=\dfrac{1}{sin10^o}\) .

10 tháng 5 2017

a) \(sin6\alpha cot3\alpha cos6\alpha=2.sin3\alpha.cos3\alpha\dfrac{cos3\alpha}{sin3\alpha}-cos6\alpha\)
\(=2cos^23\alpha-\left(2cos^23\alpha-1\right)=1\) (Không phụ thuộc vào x).

10 tháng 5 2017

b) \(\left[tan\left(90^o-\alpha\right)-cot\left(90^o+\alpha\right)\right]^2\)\(-\left[cot\left(180^o+\alpha\right)+cot\left(270^o+\alpha\right)\right]^2\)
\(=\left[cot\alpha+cot\left(90^o-\alpha\right)\right]^2\)\(-\left[cot\alpha+cot\left(90^o+\alpha\right)\right]^2\)
\(=\left[cot\alpha+tan\alpha\right]^2-\left[cot\alpha-tan\alpha\right]^2\)
\(=4tan\alpha cot\alpha=4\). (Không phụ thuộc vào \(\alpha\)).

26 tháng 4 2017

Giải bài 4 trang 155 SGK Đại Số 10 | Giải toán lớp 10

Giải bài 4 trang 155 SGK Đại Số 10 | Giải toán lớp 10