Prove that:
$(1 + c o t A)^{2} + (1 - c o t A)^{2} = 2 c o s e c^{2} A$

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Solution

$(1 + c o t A)^{2} + (1 - c o t A)^{2} = 2 c o s e c^{2} A$
$L . H . S : (1 + \frac{c o s A}{s i n A})^{2} + (1 - \frac{c o s A}{s i n A})^{2}$ $[∵ cot A = \frac{cos A}{sin A}]$
$= (\frac{s i n A + cos A}{sin A})^{2} + (\frac{s i n A - c o s A}{s i n A})^{2}$ $[∵ s i n^{2} A + c o s^{2} A = 1]$
$= \frac{s i n^{2} A + c o s^{2} A + 2 s i n A . c o s A}{s i n^{2} A} + \frac{s i n^{2} A + c o s^{2} A - 2 s i n A . c o s A}{s i n^{2} A}$
$= \frac{1 + 2 s i n A . c o s A + 1 - 2 s i n A . c o s A}{s i n^{2} A}$
$= \frac{2}{s i n^{2} A}$ $[\frac{1}{s i n A} = c o s e c A]$
$= 2 c o s e c^{2} A = R . H . S .$

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Prove that:(1+cotA)2+(1−cotA)2=2cosec2A

(1+cotA)2+(1−cotA)2=2cosec2A L.H.S:(1+cosAsinA)2+(1−cosAsinA)2 [∵cotA=cosAsinA]=(sinA+cosAsinA)2+(sinA−cosAsinA)2 [∵sin2A+cos2A=1]=sin2A+cos2A+2sinA.cosAsin2A+sin2A+cos2A−2sinA.cosAsin2A=1+2sinA.cosA+1−2sinA.cosAsin2A=2sin2A [1sinA=cosecA] =2cosec2A=R.H.S.

Prove that:
$(1 + c o t A)^{2} + (1 - c o t A)^{2} = 2 c o s e c^{2} A$