Question

Secx - sinxtanx - cosx

Answer 1

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→ Answer :  0 is Answer .

→ Step-by-step explanation :

  $\bold{secx - sinx.tanx - cosx}\\\\=>\frac{1}{cosx}-sinx.\frac{sinx}{cosx} -cosx\\\\=>\frac{1}{cosx}-\frac{sin^2x}{cosx}-cosx\\\\=>\frac{1-sin^2x}{cosx} -cosx\\\\=>\frac{cos^2x}{cosx}-cosx$

Since,

    $\bold{sin^2x+cos^2x=1 }\\\\=>cos^2x=1-sin^2x}$

$=>cosx-cosx\\\\=>0$

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