Question

if 9 sin A + 40 cos A = 40, prove that 41 cos A = 40​

Answer 1

Step-by-step explanation:

t is given that 9sinθ + 40cosθ = 41

then we have to find all the trigonometric ratios

9sinθ + 40cosθ = 41

⇒9sinθ/sinθ + 40cosθ/cosθ = 41/cosθ

⇒9 + 40tanθ = 41secθ

⇒(9 + 40tanθ)² = (41secθ)²

⇒9² + 40²tan²θ + 2(9)(40)tanθ = 41²sec²θ

⇒81 + 1600tan²θ + 720tanθ = 1681 + 1681tan²θ

⇒81tan²θ - 720tanθ + 1600 = 0

⇒(9tanθ)² - 2(9tanθ)(40) + 40² = 0

⇒(9tanθ - 40)² = 0

⇒tanθ = 40/9

so, sinθ = 40/41 ,

cosecθ = 41/40,

cosθ = 9/41 ,

secθ = 41/9,

cotθ = 9/40

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Answer 2

40 cos theta + 90 sin theta = 41

9 sin theta = 41 - 40 cos theta

squaring both the sides

81 sin square theta = 1681 + 1600 cos square theta - 2 . 40 . 41 cos theta .

81 -81 cos square theta =1681 + 1600 cos square theta- 3280 cos theta .

1681 cos square theta - 3280 cos theta + 1600 = 0 .

hope this helps you..