If -3 -2- then write the value of -1-cos 2 -1-cos 2 -

Π lt; θ lt; 3 π/2 ⇒ θ lies in the 3^rd quadrant √(1 cos 2 θ/1+cos 2 θ) = √(2 sin^2 θ/2 cos^2 θ) ⇒ sin θ/ cos θ (∵ θ lies in the 3^rd quadrant) ⇒ tan ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Multiple of an Angle

If π<θ<3 π/2,...

Question

If $π < θ < \frac{3 π}{2}$ , then write the value of

$\sqrt{\frac{1 - c o s 2 θ}{1 + c o s 2 θ}}$

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Solution

$π < θ < \frac{3 π}{2}$

$\Rightarrow θ$ lies in the $3^{r d}$ quadrant

$\sqrt{\frac{1 - c o s 2 θ}{1 + c o s 2 θ}} = \sqrt{\frac{2 s i n^{2} θ}{2 c o s^{2} θ}} \Rightarrow \frac{s i n θ}{- c o s θ} (∵ θ lies in the 3^{r d} quadrant) \Rightarrow t a n θ$

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If π<θ<3π2, then write the value of √1−cos 2θ1+cos 2θ

π<θ<3π2 ⇒ θ lies in the 3rd quadrant √1−cos 2θ1+cos 2θ=√2 sin2 θ2 cos2 θ⇒ sin θ−cos θ (∵ θ lies in the 3rd quadrant)⇒ tan θ

If $π < θ < \frac{3 π}{2}$ , then write the value of

$\sqrt{\frac{1 - c o s 2 θ}{1 + c o s 2 θ}}$

$π < θ < \frac{3 π}{2}$

$\Rightarrow θ$ lies in the $3^{r d}$ quadrant

$\sqrt{\frac{1 - c o s 2 θ}{1 + c o s 2 θ}} = \sqrt{\frac{2 s i n^{2} θ}{2 c o s^{2} θ}} \Rightarrow \frac{s i n θ}{- c o s θ} (∵ θ lies in the 3^{r d} quadrant) \Rightarrow t a n θ$