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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