\(\dfrac{1-\sin^2a\cos^2a}{\sin^2a}-\sin^2a\)
\(=\dfrac{1-\sin^2a\cos^2a-\sin^2a\sin^2a}{\sin^2a}\)
\(=\dfrac{1-\sin^2a\left(\cos^2a+\sin^2a\right)}{\sin^2a}\)
\(=\dfrac{\cos^2a+\sin^2a-\sin^2a}{\sin^2a}\)
\(=\dfrac{\cos^2a}{\sin^2a}=\cot^2a\)
\(\dfrac{1-\sin^2a\cos^2a}{\sin^2a}-\sin^2a\)
\(=\dfrac{1-\sin^2a\cos^2a-\sin^2a\sin^2a}{\sin^2a}\)
\(=\dfrac{1-\sin^2a\left(\cos^2a+\sin^2a\right)}{\sin^2a}\)
\(=\dfrac{\cos^2a+\sin^2a-\sin^2a}{\sin^2a}\)
\(=\dfrac{\cos^2a}{\sin^2a}=\cot^2a\)
Chứng minh: \(\dfrac{sin^2a-tan^2a}{cos^2a-cot^2a}\) = tan6a
Chứng minh:
\(1-cot^4a=\frac{2}{sin^2a}-\frac{1}{sin^4a}\)
Chứng minh rằng:
a) \(sin\left(a+b\right).sin\left(a-b\right)=sin^2a-sin^2b=cos^2b-cos^2a\)
b) \(4sin\left(x+\dfrac{\Pi}{3}\right).sin\left(x-\dfrac{\Pi}{3}\right)=4sin^2x-3\)
c) \(sin\left(x+\dfrac{\Pi}{4}\right)-sin\left(x-\dfrac{\Pi}{4}\right)=\sqrt{2}cosx\)
d) \(\dfrac{1}{sin10^0}-\dfrac{\sqrt{3}}{cos10^0}=4\)
Chứng minh rằng:
a) \(\left(\dfrac{tga+cosa}{1+cotga.cosa}\right)^n=\dfrac{tg^na+cos^na}{1+cotg^na.cos^na},\forall n\in Z^+\)
b) \(tga.tgb=\dfrac{tga+tgb}{cotga+cotgb}\)
c) \(\dfrac{tg^2a-tg^2b}{tg^2a.tg^2b}=\dfrac{sin^2a-sin^2b}{sin^2a.sin^2b}\)
g) \(\dfrac{1}{4}\left(\sqrt{\dfrac{1+sina}{1-sina}}-\sqrt{\dfrac{1-sina}{1+sina}}\right)^2=tg^2a\)
\(CMR:\frac{2+\sin^2a\cos^2a}{1+\cos^2a}=1+\sin^2a\)
Cm biểu thức ko phụ thuộc x
\(A=\dfrac{cot^2a-cos^2a}{cot^2a}+\dfrac{sinacosa}{cota}\)
A= sin8x+\(2cos^2x\left(4x+\dfrac{\pi}{4}\right)\)
Cm đẳng thức
\(\dfrac{sin2a-2sina}{sin2a+2sina}+tan^2\dfrac{a}{2}=0\)
\(\dfrac{sina}{1+cosa}+\dfrac{1+cosa}{sina}=\dfrac{2}{sina}\)
\(\dfrac{sin^2x}{sinx-cosx}-\dfrac{sinx+cosx}{tan^2x-1}=sinx+cosx\)
\(\dfrac{sin\left(a+b\right)sin\left(a-b\right)}{1-tan^2a.cot^2b}=-cos^2a.sin^2b\)
VỚI tam giác ABC bất kì , tìm giá trị lớn nhất của
M = \(\dfrac{\sin^2A+\sin^2B+\sin^2C}{\cos^2A+\cos^2B+\cos^2C}\)
\(CMR:\sin^6a+\cos^6a=1-3\sin^2a\cos^2a\)
mn cho e hỏi cách giải chi tiết câu này với ạ:
Chứng minh rằng tam giác ABC vuông nếu sin^2A+sin^B+sin^2C=2
em cảm ơn ạ