If -2- 3 -2- then the probability that cos-sin- is

The correct option is B 1/2cos θ gt; sin θ ⇒θϵ[ π/2, π/4 )∪(5π/4,3π/2 ] P(E)=length of above interval/length of above interval=1/2. ...

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Trigonometric Ratios of Compound Angles

If θϵ[-π/2, 3...

Question

If $θ ϵ [\frac{- π}{2}, \frac{3 π}{2}]$ , then the probability that $c o s θ > s i n θ$ is

\frac{1}{8}

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\frac{1}{4}

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\frac{1}{2}

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\frac{3}{4}

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If $m = (c o s θ - s i n θ)$ and $n = (c o s θ + s i n θ)$ then show that

$\sqrt{\frac{m}{n}} + \sqrt{\frac{n}{m}} = \frac{2}{\sqrt{1 - t a n^{2} θ}}$

t a n (\begin{matrix} \frac{π}{2} s i n θ \end{matrix}) = c o t (\begin{matrix} \frac{π}{2} c o s θ \end{matrix})

s i n θ + c o s θ = \pm \sqrt{2}

- \sqrt{2} \leq s i n θ + c o s θ \leq \sqrt{2}

θ \in [\frac{π}{2}, \frac{3 π}{2}]

2 {cos}^{2} θ + sin θ \leq 2.

m

\sqrt{\frac{1 - sin θ}{1 + sin θ}} + \sqrt{\frac{1 + sin θ}{1 - sin θ}} = \frac{- m}{cos θ}, \frac{π}{2} < θ < π

sin (θ - \frac{π}{6}) + cos (θ - \frac{π}{3}) = \sqrt{3} . sin θ

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