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Prove that :
$(\sin A - \operatorname{cosec} A)^2 + (\cos A - \sec A)^2 = \tan^2 A + \cot^2 A-1$
$(\sin A - \operatorname{cosec} A)^2 + (\cos A - \sec A)^2 = \tan^2 A + \cot^2 A-1$
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Solution (a) LHS $= \sin^2 A +\operatorname{cosec}^2 A - 2\sin A.\operatorname{cosec} A + \cos^2 A + \sec^2 A - 2\cos A.\sec A$ (1 Mark)
$= (\sin^2 A + \cos^2 A) + 1 + \cot^2 A + 1 + \tan^2 A - 2 - 2$ (1 Mark)
$= 1 + \cot^2 A + \tan^2 A +2-4$ (1/2 Mark)
$= \tan^2 A + \cot^2 A - 1 = \text{RHS}$ (1/2 Mark)
$= (\sin^2 A + \cos^2 A) + 1 + \cot^2 A + 1 + \tan^2 A - 2 - 2$ (1 Mark)
$= 1 + \cot^2 A + \tan^2 A +2-4$ (1/2 Mark)
$= \tan^2 A + \cot^2 A - 1 = \text{RHS}$ (1/2 Mark)