Question

Cos A
/1-tanA
+
SinA
/1-CotA
=
COSA + sinA

Answer 1

Answer:

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

$\sf{\dfrac{cos\:A}{1-tan\:A} +\dfrac{sin\:A}{1-cot\:A} =cos\:A+sin\:A}$

$\Large{\underline{\underline{\bf{To\:Prove:}}}}$

LHS = RHS

$\Large{\underline{\underline{\bf{Proof:}}}}$

➜ The LHS is given by

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

➜ Simplifying by using identities,

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

➜ Cross multiplying,

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

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

➜ Taking the negative outside,

     $\sf{\dfrac{cos^{2} A}{cos\:A-sin\:A}-\dfrac{sin^{2} A}{cos\:A-sin\:A} }$

        $\sf{\dfrac{cos^{2}\:A-sin^{2}\:A }{cos\:A-sin\:A}}$

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

➜ Cancelling cos A - sin A on both numerator and denominator

     $\sf{\dfrac{(cos\:A+sin\:A)\cancel{(cos\:A-sin\:A)}}{\cancel{cos\:A-sin\:A}}}$

       $\sf{cos\:A+sin\:A}$

       = RHS

➜ Hence proved.

$\Large{\underline{\underline{\bf{Identities\:used:}}}}$

➜ tan A = sin A/ cos A

➜ cot A = cos A / sin A

➜ (a + b) × (a - b) = a² - b²