If 3cos θ + 4sin θ = 5 then 3sin θ - 4cos θ is

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3.Trigonometrical Ratios, Functions and Identities

normal

If $3\cos \theta + 4\sin \theta = 5$ then $3\sin \theta - 4\cos \theta $ is

A

$1$

B

$-1$

C

$0$

D

$\frac {1}{2}$

Solution

$3\cos \theta + 4\sin \theta = 5$ and $3\sin \theta – 4\cos \theta = x$

Square and add $x^2 = 0$

Standard 11

Mathematics

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