Question

sinA(secA+cosecA) - cosA(secA-cosecA)=secAcosecA

Answer 1

LHS

$sin A(sec A + cosec A ) - cos A(sec A - cosec A) = sec A cosec A$

$sinA .(\frac{1}{cos A} + \frac{1}{sin A}) - cos A.(\frac{1}{cos A} - \frac{1}{sin A}) \\\\ sin A(\frac{sin A+cos A}{sin A cos A} )+ cosA(\frac{sin A- cos A}{cos A sin A} ) \\\\ \frac{sinA+ cos A}{ cos A } + \frac{ sin A - cos A}{sin A}\\\\ \frac{ sinA(sinA+cosA)-cos A(sin A - cosA)}{cos A. sin A} \\\\ \frac{ sin^{2} A + sin A cos A- sin AcosA + cos^{2} A}{cos A sin A} \\\\ \frac{sin^{2} A+ cos^{2} A} {sinA. cos A} \\\\ \frac{1}{sin A.cos A } \\\\ sec\: A.cosec\: A$

RHS

LHS = RHS

Hence Proved

For whole answer refer attachment

Thanks

Answer 2

Hi ,

LHS=sinA(secA+cosecA)-cosA(secA-cosecA)

= sinAsecA + sinAcosecA

-cosAsecA+cosAcosecA

= sinA/cosA + 1 - 1 + cosA/sinA

= sinA/cosA + cosA/sinA

= ( sin²A + cos²A )/cosAsinA

= 1/cosAsinA

= secAcosecA

= RHS

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