Answer:
Step-by-step explanation:
[tex](sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA[/tex]
we take the LHS so here goes,
[tex](sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA\\(\frac{1}{cosA} -\frac{1}{sinA})(1+\frac{sinA}{cosA}+\frac{cosA}{sinA})\\\\(\frac{sinA-cosA}{sinAcosA})(\frac{sinAcosA+sin^2A+cos^2A}{sinAcosA})\\[/tex]
since , [tex]sin^2A+cos^2A=1[/tex]
the identity becomes,
[tex](\frac{sinA-cosA}{sinAcosA})(\frac{1+sinAcosA}{sinAcosA})\\\\(\frac{sinA+sin^2AcosA-cosA-cos^2AsinA}{sin^2Acos^2A})\\\\[/tex]
now, we know,
[tex]sin^2A=1-cos^2A[/tex] and [tex]cos^2A=1-sin^2A[/tex]
the identity becomes,
[tex](\frac{sinA+(1-cos^2A)cosA-cosA-(1-sin^2A)sinA}{sin^2Acos^2A} )\\\\[/tex]
[tex](\frac{sinA+cosA-cos^3A-cosA-sinA+sin^3A}{sin^2Acos^2A})[/tex]
sin A and cos A cancel out it becomes zero
[tex]\frac{sin^3A-cos^3A}{sin^2Acos^2A} \\\\[/tex]
Splitting the denominator the identity becomes
[tex]\frac{sin^3A}{sin^2Acos^2A}-\frac{cos^3A}{sin^2Acos^2A} \\\\\frac{sinA}{cos^2A} - \frac{cosA}{sin^2A} \\\\\frac{sinA}{cosA}(\frac{1}{cosA})-\frac{cosA}{sinA}(\frac{1}{sinA})\\\\[/tex]
Hence,
[tex]tanAsecA-cotAcosecA[/tex]
