Question

Prove that (1+sec A+tan A)(1-cosec A+cot A)=2​

Answer 1

Answer:

Taking the LHS we first simply multiply the two brackets and get the result as follows:

1 + tan A + sec A + cot A +1 + cosec A - cosec A - sec A - 1/sinAcosA :

then the next step is to cancel the terms with the positive and negative sign and we are left with:

2 + tan A + cot A -1/sinAcosA :

then the third step is to convert the above equation into sin-cos form as follows:

2+ sinA/cosA + cosA/sinA - 1/sinAcosA :

the fourth is step is to take the LCM of the sin -cos terms leaving 2 as it is and we get:

2 + {sin[sqaure]A + cos[sqaure]A - 1}/sinAcosA :

the last and the final step is to apply the identity sin[square]A + cos[square]A=1 and then cancelling 1 from 1 we are left with 2.

hence LHS = RHS

Answer 2

Answer:

Taking the LHS we first simply multiply the two brackets and get the result as follows:

1 + tan A + sec A + cot A +1 + cosec A - cosec A - sec A - 1/sinAcosA :

then the next step is to cancel the terms with the positive and negative sign and we are left with:

2 + tan A + cot A -1/sinAcosA :

then the third step is to convert the above equation into sin-cos form as follows:

2+ sinA/cosA + cosA/sinA - 1/sinAcosA :

the fourth is step is to take the LCM of the sin -cos terms leaving 2 as it is and we get:

2 + {sin[sqaure]A + cos[sqaure]A - 1}/sinAcosA :

the last and the final step is to apply the identity sin[square]A + cos[square]A=1 and then cancelling 1 from 1 we are left with 2.

hence LHS = RHS