If - x- 2- then - 1-cos x - 1-cos x - 1-cos x - 1-cos x -

Let y=√(1+cos x)+√(1 cos x)√(1+cos x) √(1 cos x) ∴ y=√(2)cosx2+√(2)sinx2√(2)cosx2 √(2)sinx2 ∴ y=1+tanx21 tanx2 ∴ #160; y=tan(π4+x2) ...

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Standard XII

Mathematics

Domain and Range of Trigonometric Ratios

If π< x< 2π...

Question

If $π < x < 2 π$ then $\frac{\sqrt{1 + cos x} + \sqrt{1 - cos x}}{\sqrt{1 + cos x} - \sqrt{1 - cos x}} =$

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Solution

let
$y = \frac{\sqrt{1 + cos x} + \sqrt{1 - cos x}}{\sqrt{1 + cos x} - \sqrt{1 - cos x}}$

$∴ y = \frac{\sqrt{2} cos \frac{x}{2} + \sqrt{2} sin \frac{x}{2}}{\sqrt{2} cos \frac{x}{2} - \sqrt{2} sin \frac{x}{2}}$

$∴ y = \frac{1 + tan \frac{x}{2}}{1 - tan \frac{x}{2}}$

$∴ y = tan (\frac{π}{4} + \frac{x}{2})$

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Similar questions

Q. If

π < x < 2 π

, prove that

\frac{\sqrt{1 + cos x} + \sqrt{1 - cos x}}{\sqrt{1 + cos x} - \sqrt{1 - cos x}} = cot (\frac{x}{b} + \frac{π}{4})

.Find

b

If x ∈ ( $π$ , 2 $π$ ) and $\frac{\sqrt{1 + c o s x} + \sqrt{1 - c o s x}}{\sqrt{1 + c o s x} - \sqrt{1 - c o s x}}$ = $c o t (a + \frac{x}{2})$ , then a is equal to

If $\frac{\sqrt{1 + c o s x} + \sqrt{1 - c o s x}}{\sqrt{1 + c o s x} = \sqrt{1 - c o s x}} = c o t (a + \frac{x}{2}), x ϵ (π, 2 π)$ then a___

Q. Prove that

{tan}^{- 1} (\frac{\sqrt{1 + cos x} + \sqrt{1 - cos x}}{\sqrt{1 + cos x} - \sqrt{1 - cos x}}) = \frac{π}{4} - \frac{x}{2}

π < x < \frac{3 π}{2}

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If π<x<2π then √1+cosx+√1−cosx√1+cosx−√1−cosx=

lety=√1+cosx+√1−cosx√1+cosx−√1−cosx∴y=√2cosx2+√2sinx2√2cosx2−√2sinx2∴y=1+tanx21−tanx2∴y=tan(π4+x2)

If $π < x < 2 π$ then $\frac{\sqrt{1 + cos x} + \sqrt{1 - cos x}}{\sqrt{1 + cos x} - \sqrt{1 - cos x}} =$