Question

chứng minh rằng:
a)\(1-\frac{sin^2a}{1+c\text{os}a}-\frac{c\text{os}^2a}{1+t\text{ana}}=sin\text{a}.c\text{os}a\)
b)\(\frac{c\text{os}a.cot\text{a}-sin\text{a}.t\text{ana}}{\frac{1}{sin\text{a}}-\frac{1}{c\text{os}a}}=1+sin\text{a}.c\text{os}a\)

Answer 1

Lời giải:

a) Sửa đề:

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

\(=1-\frac{\sin ^3a}{\sin a+\cos a}-\frac{\cos ^3a}{\sin a+\cos a}=1-\frac{\sin ^3a+\cos ^3a}{\sin a+\cos a}=1-\frac{(\sin a+\cos a)(\sin ^2a-\sin a\cos a+\cos ^2a)}{\sin a+\cos a}\)

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

\(=\sin a\cos a\) (đpcm)

b)

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

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

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

(đpcm)

Answer 2

Lời giải:

a) Sửa đề:

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

\(=1-\frac{\sin ^3a}{\sin a+\cos a}-\frac{\cos ^3a}{\sin a+\cos a}=1-\frac{\sin ^3a+\cos ^3a}{\sin a+\cos a}=1-\frac{(\sin a+\cos a)(\sin ^2a-\sin a\cos a+\cos ^2a)}{\sin a+\cos a}\)

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

\(=\sin a\cos a\) (đpcm)

b)

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

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

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

(đpcm)

Answer 3

Lời giải:

a) Sửa đề:

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

\(=1-\frac{\sin ^3a}{\sin a+\cos a}-\frac{\cos ^3a}{\sin a+\cos a}=1-\frac{\sin ^3a+\cos ^3a}{\sin a+\cos a}=1-\frac{(\sin a+\cos a)(\sin ^2a-\sin a\cos a+\cos ^2a)}{\sin a+\cos a}\)

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

\(=\sin a\cos a\) (đpcm)

b)

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

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

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

(đpcm)

Answer 4

Lời giải:

a) Sửa đề:

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

\(=1-\frac{\sin ^3a}{\sin a+\cos a}-\frac{\cos ^3a}{\sin a+\cos a}=1-\frac{\sin ^3a+\cos ^3a}{\sin a+\cos a}=1-\frac{(\sin a+\cos a)(\sin ^2a-\sin a\cos a+\cos ^2a)}{\sin a+\cos a}\)

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

\(=\sin a\cos a\) (đpcm)

b)

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

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

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

(đpcm)