Question

(1+cota+seca)(1+cota-seca)=2cota

Answer 1

Step-by-step explanation:

Taking RHS

Taking RHS(1 cotA)/secA = sin^2A/(1-cosA)

Taking RHS(1 cotA)/secA = sin^2A/(1-cosA)Sin A ^ 2 = 1 – cos A ^2

Taking RHS(1 cotA)/secA = sin^2A/(1-cosA)Sin A ^ 2 = 1 – cos A ^2Hence, 1 – cos A ^2//(1-cosA)

Taking RHS(1 cotA)/secA = sin^2A/(1-cosA)Sin A ^ 2 = 1 – cos A ^2Hence, 1 – cos A ^2//(1-cosA)A^2 – B^2 = (A – B) (A + B)

Taking RHS(1 cotA)/secA = sin^2A/(1-cosA)Sin A ^ 2 = 1 – cos A ^2Hence, 1 – cos A ^2//(1-cosA)A^2 – B^2 = (A – B) (A + B)(1 - cosA) (1 + cosA) /(1-cosA)

Taking RHS(1 cotA)/secA = sin^2A/(1-cosA)Sin A ^ 2 = 1 – cos A ^2Hence, 1 – cos A ^2//(1-cosA)A^2 – B^2 = (A – B) (A + B)(1 - cosA) (1 + cosA) /(1-cosA)(1+ cosA)

Taking RHS(1 cotA)/secA = sin^2A/(1-cosA)Sin A ^ 2 = 1 – cos A ^2Hence, 1 – cos A ^2//(1-cosA)A^2 – B^2 = (A – B) (A + B)(1 - cosA) (1 + cosA) /(1-cosA)(1+ cosA)Cos A = 1/sec A

Taking RHS(1 cotA)/secA = sin^2A/(1-cosA)Sin A ^ 2 = 1 – cos A ^2Hence, 1 – cos A ^2//(1-cosA)A^2 – B^2 = (A – B) (A + B)(1 - cosA) (1 + cosA) /(1-cosA)(1+ cosA)Cos A = 1/sec A1 + 1/secA

Taking RHS(1 cotA)/secA = sin^2A/(1-cosA)Sin A ^ 2 = 1 – cos A ^2Hence, 1 – cos A ^2//(1-cosA)A^2 – B^2 = (A – B) (A + B)(1 - cosA) (1 + cosA) /(1-cosA)(1+ cosA)Cos A = 1/sec A1 + 1/secAsecA + 1/secA = RHS

Taking RHS(1 cotA)/secA = sin^2A/(1-cosA)Sin A ^ 2 = 1 – cos A ^2Hence, 1 – cos A ^2//(1-cosA)A^2 – B^2 = (A – B) (A + B)(1 - cosA) (1 + cosA) /(1-cosA)(1+ cosA)Cos A = 1/sec A1 + 1/secAsecA + 1/secA = RHSthe LHS part of the question doesn’t make sense.

Answer 2

Answer:

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