Saved Bookmarks
| 1. |
move that sne-cose-1-secoltare , using the identity sec28-Prove that |
|
Answer» LHS = (sinθ - cosθ + 1)/(sinθ + cosθ - 1) dividing by cosθ both Numerator and denominator = (sinθ/cosθ - cosθ/cosθ + 1/cosθ)/(sinθ/cosθ + cosθ/cosθ - 1/cosθ)= (tanθ + secθ - 1)/(tanθ - secθ + 1) Multiply (tanθ - secθ) with both Numerator and denominator = (tanθ + secθ - 1)(tanθ - secθ)/(tanθ - secθ + 1)(tanθ - secθ)= {(tan²θ - sec²θ) - (tanθ - secθ)}/(tanθ - secθ + 1)(tanθ - secθ)= (-1 - tanθ + secθ)/(tanθ - secθ + 1)(tanθ - secθ) [ ∵ sec²x - tan²x = 1 ]= -1/(tanθ - secθ) = 1/(secθ - tanθ) = RHS Please like the solution 👍 ✔️👍 |
|