Phương pháp giải:

Sử dụng công thức lượng giác: $\left\{ \begin{array}{l}{\sin ^2}\alpha  + {\cos ^2}\alpha  = 1\\\tan \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }}\\\cot \alpha  = \frac{{\cos \alpha }}{{\sin \alpha }}\end{array} \right..$

Sử dụng hằng đẳng thức.

Lời giải chi tiết:

a) $1 + {\rm{ }}{\tan ^2}x{\rm{ }} = \frac{1}{{{{\cos }^2}x}}$

$VT = 1 + {\rm{ }}{\tan ^2}x$$= 1 + \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}$$= \frac{{{{\cos }^2}x + {{\sin }^2}x}}{{{{\cos }^2}x}}$$= \frac{1}{{{{\cos }^2}x}} = VP$(đpcm)

b) $1 + {\cot ^2}x{\rm{ }} = \frac{1}{{{{\sin }^2}x}}$

$VT = 1 + {\cot ^2}x{\rm{ }} = 1 + \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}}$$= \frac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\sin }^2}x}} = \frac{1}{{{{\sin }^2}x}}$ (đpcm)

c) ${\cos ^4}x--{\sin ^4}x = 2{\cos ^2}x{\rm{ }} - 1$

$\begin{array}{l}{\cos ^4}x--{\sin ^4}x = \left( {{{\cos }^2}x--{{\sin }^2}x} \right)\left( {{{\cos }^2}x{\rm{  +  }}{{\sin }^2}x} \right)\\ = {\cos ^2}x--{\sin ^2}x = {\cos ^2}x - \left( {1 - {{\cos }^2}x{\rm{ }}} \right)\\ = 2{\cos ^2}x{\rm{ }} - 1\,\,\,\,\left( {dpcm} \right)\end{array}$

d. ${\sin ^6}x + {\cos ^6}x{\rm{ }} = 1 - 3{\sin ^2}x.{\cos ^2}x$

$\begin{array}{l}{\sin ^6}x + {\cos ^6}x = \left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^4}x - {{\sin }^2}x.{{\cos }^2}x + {{\cos }^4}x} \right)\\ = \left( {{{\sin }^4}x + {{\cos }^4}x} \right) - {\sin ^2}x.{\cos ^2}x\\ = {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x.{\cos ^2}x - {\sin ^2}x.{\cos ^2}x\\ = 1 - 3{\sin ^2}x.{\cos ^2}x\,\,\,\,\left( {dpcm} \right)\end{array}$