The maximum value of $$35sin$$⁡x−$$12cos$$⁡$$x+19$$ is

The maximum value of $$35sin$$⁡x−$$12cos$$⁡$$x+19$$ isequal to $$3minimum$$ value of $$5sin$$⁡x−$$12cos$$⁡$$x+19$$ $$=319-25+144=319-13=36=12$$

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The maximum value of $\frac{3}{5 \sin x - 12 \cos x + 19}$ is

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\frac{1}{2}

\frac{1}{3}

\frac{1}{4}

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Correct option is B

The maximum value of $\frac{3}{5 \sin x - 12 \cos x + 19}$ is

equal to $\frac{3}{m i n i m u m v a l u e o f (5 \sin x - 12 \cos x + 19)} = \frac{3}{19 - \sqrt{25 + 144}} = \frac{3}{19 - 13} = \frac{3}{6} = \frac{1}{2}$

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The maximum value of 35sin⁡x−12cos⁡x+19 is

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The maximum value of $\frac{3}{5 \sin x - 12 \cos x + 19}$ is