Đáp án:

a)  $A=4$

b)  $B=2-{{\cot }^{2}}x$

 

Giải thích các bước giải:

a)

$A={{\left( \tan x+\cot x \right)}^{2}}-{{\left( \tan x-\cot x \right)}^{2}}$

$A={{\tan }^{2}}x+2\tan x.\cot x+{{\cot }^{2}}x-\left( {{\tan }^{2}}x-2\tan x.\cot x+{{\cot }^{2}}x \right)$

$A=4\tan x.\cot x$

$A=4$

 

b)

$B={{\cos }^{2}}x.{{\cot }^{2}}x+2{{\sin }^{2}}x-2{{\cot }^{2}}x+3{{\cos }^{2}}x$

$B=\left( {{\cos }^{2}}x.{{\cot }^{2}}x-{{\cot }^{2}}x+{{\cos }^{2}}x \right)+\left( 2{{\sin }^{2}}x+2{{\cos }^{2}}x \right)-{{\cot }^{2}}x$

$B=\left( {{\cos }^{2}}x\cdot \dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x}-\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x}+{{\cos }^{2}}x \right)+2\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)-{{\cot }^{2}}x$

$B=\dfrac{{{\cos }^{4}}x-{{\cos }^{2}}x+{{\cos }^{2}}x.{{\sin }^{2}}x}{{{\sin }^{2}}x}+2-{{\cot }^{2}}x$

$B=\dfrac{{{\cos }^{4}}x-{{\cos }^{2}}x\left( 1-{{\sin }^{2}}x \right)}{{{\sin }^{2}}x}+2-{{\cot }^{2}}x$

$B=\dfrac{{{\cos }^{4}}x-{{\cos }^{2}}x.{{\cos }^{2}}x}{{{\sin }^{2}}x}+2-{{\cot }^{2}}x$

$B=\dfrac{0}{{{\sin }^{2}}x}+2-{{\cot }^{2}}x$

$B=2-{{\cot }^{2}}x$