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Ratling [72]
2 years ago
13

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

Mathematics
1 answer:
mash [69]2 years ago
3 0

Answer:

Step-by-step explanation:

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

we take the LHS so here goes,

(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})\\

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

the identity becomes,

(\frac{sinA-cosA}{sinAcosA})(\frac{1+sinAcosA}{sinAcosA})\\\\(\frac{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,

(\frac{sinA+(1-cos^2A)cosA-cosA-(1-sin^2A)sinA}{sin^2Acos^2A} )\\\\

(\frac{sinA+cosA-cos^3A-cosA-sinA+sin^3A}{sin^2Acos^2A})

sin A and cos A cancel out it becomes zero

\frac{sin^3A-cos^3A}{sin^2Acos^2A} \\\\

Splitting the denominator the identity becomes

\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})\\\\

Hence,

tanAsecA-cotAcosecA

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