Question

Prove that: (Sec A- cosec A)(1+ tan A+cot A) = tan A× sec A - cot A × cosec A

Answer 1

Explanation:

`(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA`

we take the LHS so here goes,

`(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA\\((1)/(cosA) -(1)/(sinA))(1+(sinA)/(cosA)+(cosA)/(sinA))\\\\((sinA-cosA)/(sinAcosA))((sinAcosA+sin^2A+cos^2A)/(sinAcosA))\\`

since ,
`sin^2A+cos^2A=1`

the identity becomes,

`((sinA-cosA)/(sinAcosA))((1+sinAcosA)/(sinAcosA))\\\\((sinA+sin^2AcosA-cosA-cos^2AsinA)/(sin^2Acos^2A))\\\\`

now, we know,

`sin^2A=1-cos^2A` and
`cos^2A=1-sin^2A`

the identity becomes,

`((sinA+(1-cos^2A)cosA-cosA-(1-sin^2A)sinA)/(sin^2Acos^2A) )\\\\`

`((sinA+cosA-cos^3A-cosA-sinA+sin^3A)/(sin^2Acos^2A))`

sin A and cos A cancel out it becomes zero

`(sin^3A-cos^3A)/(sin^2Acos^2A) \\\\`

Splitting the denominator the identity becomes

`(sin^3A)/(sin^2Acos^2A)-(cos^3A)/(sin^2Acos^2A) \\\\(sinA)/(cos^2A) - (cosA)/(sin^2A) \\\\(sinA)/(cosA)((1)/(cosA))-(cosA)/(sinA)((1)/(sinA))\\\\`

Hence,

`tanAsecA-cotAcosecA`