(1+tan+sec)(1+cot-cos)=
Answers
Taking the LHS we first simply multiply the two brackets and get the result as follows:
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 :
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:
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 :
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:
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 :
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:
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 :
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.
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
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