Question

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

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is A 0

Let $4sin\theta -3cos\theta =a$

On squaring both sides, we get

$16si{n}^2\theta +9co{s}^2\theta -24sin\theta cos\theta =a^2$....(i)

Also, $3sin\theta +4cos\theta =5$ [Given]

Again, on squaring both sides, we get

$9si{n}^2\theta +16co{s}^2\theta +24sin\theta cos\theta =25$.......(ii)

Adding (i) and (ii) we get

$9+16=a^2+25$

$\Rightarrow 25=a^2+25$

$\Rightarrow a^2=0$

$\Rightarrow a=0$