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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