If tan theta =20/21 , evaluate 1-sin theta +cos theta /1 +sin theta +cos theta.
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HI !
tanθ = 20/21 = opposite side(o)/adjacent side(a)
hypotenuse = h = o² + a²
Let , o = 20k and a = 21k
h² = (20k)² + (21k)²
= 841k²
h = 29 k
sinθ = o/h
= 20k/29k
=20/29
cosθ = a/h
= 21k/29k
= 21/29
1 - sinθ + cosθ ÷ 1 + sinθ + cosθ
= {1 - 20/29 + 21/29 / 1 } ÷ {1 + 20/29 + 21/29}
= {29/29 - 20/29 + 21/29) ÷ {29/29+20/29+21/29}
= {29 - 20 + 21/29} ÷ {29+20+21/29}
= 30/29 ÷ 70/29
29 gets cancelled on both denominator and numerator.
= 30/70
= 3/7
tanθ = 20/21 = opposite side(o)/adjacent side(a)
hypotenuse = h = o² + a²
Let , o = 20k and a = 21k
h² = (20k)² + (21k)²
= 841k²
h = 29 k
sinθ = o/h
= 20k/29k
=20/29
cosθ = a/h
= 21k/29k
= 21/29
1 - sinθ + cosθ ÷ 1 + sinθ + cosθ
= {1 - 20/29 + 21/29 / 1 } ÷ {1 + 20/29 + 21/29}
= {29/29 - 20/29 + 21/29) ÷ {29/29+20/29+21/29}
= {29 - 20 + 21/29} ÷ {29+20+21/29}
= 30/29 ÷ 70/29
29 gets cancelled on both denominator and numerator.
= 30/70
= 3/7
Answered by
1
hope it help you .
@rajukumar ◆◆◆◆sorry for wrong answer.
@rajukumar ◆◆◆◆sorry for wrong answer.
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