Question

Please help me as early as you can.....Prove that LHS=RHS​

Answer 1

Step-by-step explanation:

• Solving both L.H.S and R.H.S

$\sec( \alpha ) = \frac{1}{ \cos( \alpha ) } \\ \\ \tan( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) }$

• Sin²x + cos²x = 1

cos²x = 1 - sin²x