Question

SinA( 1+ tan A ) + cos A ( 1 + cot A ) = sec A + cosec A

Answer 1

Sin A ( 1 + tan A) + cos A ( 1 + cot A) = Sec A + cosec A

Solving the LHS

Solving the brackets

→ Sin A + Sin A tan A + cos A + cos A cot A

Now , we know that

• Tan A = Sin A/ Cos A
• Cot A = Cos A / sin A

•°• Sin A + (Sin A tan A) + cos A + (cos A cot A)

→ Sin A + (Sin A . Sin A/ cos A) + cos A +( cos A. cos A/ sin A )

→ Sin A + (sin²A / cos A) + cos A + (cos² A / sin A )

Taking the fractions :

$\sf \dfrac{ {sin}^{2}A \: cosA + {sin}^{3} A + {cos}^{2}A \: sin \: A+ {cos}^{3}A}{sin \: A \: cos \: A}$

$\sf \dfrac{{sin}^{2} A(cos \: A \: + sin \: A) + {cos}^{2}A(sin \: A + cos \: A)}{cos \: A\ sin\ A}$

$\sf \dfrac{( {sin}^{2}A + {cos}^{2} A) + (sin \: A + cos \: A) }{cos \: A \: sin \: A}$

$\sf \dfrac{sin \: A \: + cos \: A}{cos \: A \: sin \: A}$

Separating the terms :

$\sf \dfrac{sin \: A}{cos \: A \: sin \: A} + \dfrac{cos \: A}{cos \: A \: sin \: A}$

$\sf \dfrac{1}{cos \: A} + \dfrac{1}{sin \: A}$

$\implies \boxed{\sf sec \: A + cosec \: A}$

LHS = RHS

Hence proved