Question:
SIR ACTUALLY IN TRIGONOMETRY PART 13 (EXAMPLES)...WILL YOU PLEASE EXPLAIN ME WHAT STEP YOU DID AT 8:28MINUTES AND HOW YOU GOT THE ANSWER AFTER THAT
Answer:
In the video, we have to prove that
(1 + secA)/secA = sin2 A/(1 - cosA)
LHS
(1 + secA)/secA
= {1 + {1/cosA)}/(1/cosA) (since secA = 1/cosA)
= {(cosA + 1)/cosA)}/(1/cosA)
= {(cosA + 1)*cosA}/cosA
=> 1+ cosA
Now, multiply and divdie by (1 - cosA)
=> {(1 + cosA)*(1 - cosA)}/(1 - cosA)
=> (1 - cos2 A)/(1 - cosA)
=> sin2 A/(1 - cosA) (since sin2 A + cos2 A = 1 => sin2 A = 1 - cos2 A)
= RHS
So (1 + secA)/secA = sin2 A/(1 - cosA)
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