Question

Prove that (1 +sec A) /sec A = sin2A /( 1 – cosA).​

Answer 1

Answer:

Step-by-step explanation:

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

$\dfrac{1+sec\:A}{sec\:A} =\dfrac{sin^{2}\:A }{1-cos\:A}$

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

LHS = RHS

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

→ Consider the LHS of the equation,

   $\sf {LHS = \dfrac{1+sec\:A}{sec\:A} }$

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

→ Cross multiplying,

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

→ Cancelling cos A

         =$\sf{cos\:A+1---(1)}$

→ Now consider the RHS of the equation,

  $\sf{RHS=\dfrac{sin^{2}\:A }{1-cos\:A}}$

         $\sf{=\dfrac{1-cos^{2}\:A }{1-cos\:A} }$

        $\sf{=\dfrac{(1+cos\:A)(1-cos\:A)}{1-cos\:A}}$

→ Cancelling 1 - cos A

       $\sf{=1+cos\:A---(2)}$

→ From equation 1 and 2 we see that,

  LHS = 1 + cos A

  RHS = 1 +  cos A

→ Hence LHS = RHS

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

→ sec A = 1/cos A

→ sin² A = 1 - cos² A

→ a² - b² = ( a+ b) ( a- b )