Question

prove that
$tan^{2}A+cot^{2}A=sec^{2}Acosec^{2}A-2$

Answer 1

[tex] sin^{2} A / cos ^{2} A + cos ^{2} A / sin ^{2} A (sin ^{4} A + cos ^{4} A ) / sin^{2} A cos ^{2} A [ (sin^{2} A + cos^{2} A )^{2} - 2 cos^{2} A sin^{2} A ] / cos ^{2} A sin^{2} A [ 1 - 2 cos^{2} A sin^{2} A ] / cos^{2} A sin ^{2} A 1/ cos^{2} A sin ^{2} A - 2 sec^{2} A cosec^{2} A - 2 [/tex]

Answer 2

$A=x\\\\tan^2x+cot^2x=sec^2xcosec^2x-2\\\\R=\frac{1}{cos^2x}\times\frac{1}{sin^2x}-2=\frac{1}{sin^2xcos^2x}-\frac{2sin^2xcos^2x}{sin^2xcos^2x}\\\\=\frac{1-2sin^2xcos^2x}{sin^2cos^2x}=\frac{sin^2x+cos^2x-sin^2xcos^2x-sin^2xcos^2x}{sin^2xcos^2x}\\\\=\frac{sin^2x-sin^2xcos^2x}{sin^2xcos^2x}+\frac{cos^2x-sin^2xcos^2x}{sin^2xcos^2x}=\frac{sin^2x(1-cos^2x)}{sin^2xcos^2x}+\frac{cos^2x(1-sin^2x)}{sin^2xcos^2x}$

$=\frac{1-cos^2x}{cos^2x}+\frac{1-sin^2x}{sin^2x}=\frac{sin^2x}{cos^2x}+\frac{cos^2x}{sin^2x}=tan^2x+cot^2x=L$


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$Used:\\\\sin^2x+cos^2x=1\\\\sin^2x=1-cos^2x\ and\ cos^2x=1-sin^2x\\\\tanx=\frac{sinx}{cosx}\\\\cotx=\frac{cosx}{sinx}\\\\secx=\frac{1}{cosx}\\\\cosecx=\frac{1}{sinx}$