Prove the following: (sin A + cos A)2 + (sin A – cos A)2 = 2

Prove the following:

(i) (sin A + cos A)2 + (sin A – cos A)2 = 2

(ii) cot2 A – 1/sin2 A + 1 = 0

(iii) 1/(1 + tan2 A) + 1/(1 + cot2 A) = 1

Answer :

(i) (sin A + cos A)2 + (sin A – cos A)2 = 2

LHS = (sin A + cos A)2 + (sin A – cos A)2

(a + b)2 = a2 + b2 + 2ab and (a – b)2 = a2 + b2 – 2ab

= [(sin A)2 + (cos A)2 + 2 sin A cos A] + [(sin A)2 + (cos A)2 – 2 sin A cos A]

= sin2 A + cos2 A + 2 sin A cos A + sin2 A + cos2 A – 2 sin A cos A

= sin2 A + cos2 A + sin2 A + cos2 A

= 2 sin2 A + 2 cos2 A

sin2 A + cos2 A = 1

= 2 (sin2 A + cos2 A)

= 2 (1)

= 2

= RHS

Hence, LHS = RHS.

(ii) cot2 A – 1/sin2 A + 1 = 0

LHS = cot2 A – 1/sin2 A + 1

1/sin A = cosec A

= cot2 A – cosec2 A + 1

= (1 + cot2 A) – cosec2 A

1 + cot2 A = cosec2 A

= cosec2 A – cosec2 A

= 0

= RHS

Hence, LHS = RHS.

(iii) 1/(1 + tan2 A) + 1/(1 + cot2 A) = 1

LHS = 1/(1 + tan2 A) + 1/(1 + cot2 A)

sec2 A – tan2 A = 1

⇒ sec2 A = 1 + tan2 A

⇒ cosec2 A – cot2 A = 1

⇒ cosec2 A = 1 + cot2 A

= 1/sec2 A + 1/cosec2 A

1/sec A = cos A and 1/cosec A = sin A

= cos2 A + sin2 A

= 1

= RHS

Hence,

LHS = RHS.

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