Question

Prove the identities:
(i) √[1+sinA/1-sinA] = sec A + tan A
(ii) (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 = tan2A
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Answer 1

Answer:

Step-by-step explanation:

Answer 2

Question (1) :-

• Prove :- √[1+sinA/1-sinA] = sec A + tan A

Solution :-

Taking LHS

→ √[1+sinA/1-sinA]

Rationalizing The Denominator we get,

→ √[(1+sinA)/(1-sinA) * (1+sinA)/(1+sinA) ]

→ √[(1 + sinA)² / (1 - sin²A) ]

Now, using 1 - sin²A = cos²A in Denominator ,

→ √[(1 + sinA)² / (cos²A) ]

Square root Now,

→ ( 1+ sinA) / (cosA)

→ (1/cosA) + (sinA/cosA)

Now, using (1/cosA) = secA & (sinA/cosA) = tanA, we get,

→ secA + tanA . = RHS = Proved .

_________________________

Question (2) :-

• Prove :- (1+tan²A/1+cot²A) = (1-tanA/1-cotA)² = tan²A

Solution :-

Taking LHS first :-

→ (1+tan²A)/(1+cot²A)

Putting tanA = (sinA/cosA) & cotA = (cosA/sinA) we get,

→ [ { 1+(sin²A/cos²A) } / { 1 + (cos²A/sin²A) } ]

Taking LCM ,

→ [ (cos²A+sin²A)/cos²A) / (sin²A+cos²A)/sin²A ]

using (sin²A+cos²A) = 1 now,

→ (1/cos²A) / (1/sin²A)

→ (1/cos²A) * (sin²A/1)

→ (sin²A/cos²A)

→ tan²A

Now, Solving MHS :-

→ [ (1 - tanA) / (1 - cotA) ]²

using (a - b)² = a² + b² - 2ab in N & D both,

→ (1+tan²A - 2tanA) / (1+cot² - 2cotA)

putting (1+tan²A) = sec²A & (1+cot²A) = cosec²A now,

→ (sec²A - 2tanA)/(cosec²A - 2cosA/sinA)

Putting tanA = (sinA/cosA) & cotA = (cosA/sinA) we get,

→ [{sec²A - 2*(sinA/cos A)} / {(cosec²A - (2cosA/sinA)}]

putting sec²A = (1/cos²A) & cosec²A = (1/sin²A) now,

→ [ {(1/cos²A) - 2(sinA/cosA)} / { (1/sin²A) - 2(cosA/sinA) }]

→ [(1 - 2sinAcosA)/cos²A] / [(1 - 2cosAsinA)sin²A]

→ [ { (1 - 2sinAcosA)/cos²A } * {sin²A/(1 - 2sinAcosA) } ]

→ (sin²A/cos²A)

→ tan²A

→ = R.H.S = tan²A

Hence, LHS = MHS = RHS = Proved.