A=(tanx + cotx)^2 – (Tanx + cotx)^2 B=(sinx – cosx)^2 +(sinx + cosx)^2 C=sin^2x . Tanx + cos^2x . Cotx + 2sinx.cotx

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1. Giải thích các bước giải:

    Ta có:

   \(\begin{array}{l}
   A = {\left( {\tan x + {\mathop{\rm cotx}\nolimits} } \right)^2} – {\left( {\tan x + {\mathop{\rm cotx}\nolimits} } \right)^2} = 0\\
   B = {\left( {\sin x – {\mathop{\rm cosx}\nolimits} } \right)^2} + {\left( {\sin x + \cos x} \right)^2}\\
    = {\sin ^2}x – 2\sin x.\cos x + {\cos ^2}x + {\sin ^2}x + 2\sin x\cos x + {\cos ^2}x\\
    = 2\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\\
    = 2.1 = 2\\
   C = {\sin ^2}x.\tan x + {\cos ^2}x.{\mathop{\rm cotx}\nolimits}  + 2sinx.cosx\\
    = \frac{{{{\sin }^3}x}}{{\cos x}} + \frac{{{{\cos }^3}x}}{{\sin x}} + 2sinx.cosx\\
    = \frac{{{{\sin }^4}x + {{\cos }^4}x}}{{\sin x\cos x}} + 2\sin x.\cos x\\
    = \frac{{{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^2} – 2{{\sin }^2}x.{{\cos }^2}x}}{{\sin x\cos x}} + 2\sin x\cos x\\
    = \frac{1}{{\sin x\cos x}} – 2\sin x\cos x + 2\sin x\cos x\\
    = \frac{1}{{\sin x\cos x}}
   \end{array}\)

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