Giá trị biểu thức $B=\dfrac{{{\cos }^{2}}x-{{\sin }^{2}}y}{{{\sin }^{2}}x.{{\sin }^{2}}y}-{{\cot }^{2}}x.{{\cot }^{2}}y$ bằng

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Giá trị biểu thức $B=\dfrac{{{\cos }^{2}}x-{{\sin }^{2}}y}{{{\sin }^{2}}x.{{\sin }^{2}}y}-{{\cot }^{2}}x.{{\cot }^{2}}y$ bằng

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1 năm 2021-10-13T18:07:26+00:00 2 Answers 4 views 0

1. Ta có $B=\dfrac{{{\cos }^{2}}x-{{\sin }^{2}}y}{{{\sin }^{2}}x.{{\sin }^{2}}y}-{{\cot }^{2}}x.{{\cot }^{2}}y=\dfrac{{{\cos }^{2}}x-{{\sin }^{2}}y}{{{\sin }^{2}}x{{\sin }^{2}}y}-\dfrac{{{\cos }^{2}}x.{{\cos }^{2}}y}{{{\sin }^{2}}x.{{\sin }^{2}}y}$

$=\dfrac{{{\cos }^{2}}x\left( 1-{{\cos }^{2}}y \right)-{{\sin }^{2}}y}{{{\sin }^{2}}x{{\sin }^{2}}y}=\dfrac{{{\cos }^{2}}x{{\sin }^{2}}y-{{\sin }^{2}}y}{{{\sin }^{2}}x{{\sin }^{2}}y}=\dfrac{{{\sin }^{2}}y\left( {{\cos }^{2}}x-1 \right)}{\left( 1-{{\cos }^{2}}x \right){{\sin }^{2}}y}=-1$ .

2. B=$\frac{cot²x}{sin²y}$ -$\frac{1}{sin²x}$ -cot²x.cot²y

=cot²x($\frac{1}{sin²y}$ -cot²y) -$\frac{1}{sin²x}$

=cot²x($\frac{1}{sin²y}$-$\frac{cos²y}{sin²y}$ ) -$\frac{1}{sin²x}$

=cot²x($\frac{1-cos²y}{sin²y}$ )-$\frac{1}{sin²x}$

= cot²x-$\frac{1}{sin²x}$=$\frac{cos²x}{sin²x}$  -$\frac{1}{sin²x}$=$\frac{-sin²x}{sin²x}$ =-1