Question

9sinA+40cosA=41 then prove that 41cosA=40

Answer 1

$9 \sin \alpha + 40\cos \alpha = 41 \\ \frac{9 \sin \alpha }{41} + \frac{40 \cos \alpha }{41} = 1$
we know Sin^2@ + Cos^2@ = 1


so
9/41 = Sin@
&
40/41 = Cos@

so

41 Cos@= 41 × 40/41= 40

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