Question

cosec a upon cosec a - 1 + cosec a upon cosec A + 1 is equal to 2 + 2 tan
square A

Answer 1

Is cosec A + 1/ cosec A - 1 equal to (secA+tanA) ^2?

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4 ANSWERS



Ansh Singh

Answered Jul 10

We need to prove that ,

(cosecA+1)\(cosecA−1)=(secA+tanA)2(cosecA+1)\(cosecA−1)=(secA+tanA)2

L.H.S

cosec A = 1 / sin A

cosecA+1=>1/sinA+1cosecA+1=>1/sinA+1

=> (1+sinA)/sinA−−−(i)(1+sinA)/sinA−−−(i)

cosecA−1=>1/sinA−1cosecA−1=>1/sinA−1

=> (1−sinA)/sinA−−−(ii)(1−sinA)/sinA−−−(ii)

Divide (i) and (ii)

[(1+sinA)/sinA]/[(1−sinA)/sinA][(1+sinA)/sinA]/[(1−sinA)/sinA]

(1+sinA)/(1−sinA)(1+sinA)/(1−sinA)

Multiplying by ( 1+sin A ) in numerator and denominator

[(1+sinA)2]/(1−sin2A)[(1+sinA)2]/(1−sin2A)

1 - sin^2 A = cos ^ 2 A = ( cos A ) ^ 2

(1+sinA)2/(cosA)2(1+sinA)2/(cosA)2

a^m ÷ b^m = ( a ÷ b ) ^m

(1+sinA/cosA)2(1+sinA/cosA)2

1 / cos A = sec A

sin A / cos A = tan A

(secA+tanA)2(secA+tanA)2

L.H.S = R.H.S

Answer 2

Answer:

Step-by-step explanation:

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