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

√(1+ cos x) + √(1 cos x)√(1 + cos x) √(1 cos x) = (√(1+cos x) + √(1 cos x))^2(√(1+cos x))^2 (√(1 cos x))^2 =1+cos x + 1 cos x + 2√(1+cos x)√(1 ...

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Standard XII

Mathematics

Algebra of Derivatives

If π < x < ...

Question

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

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Solution

$\frac{\sqrt{1 + cos x} + \sqrt{1 - cos x}}{\sqrt{1 + cos x} - \sqrt{1 - cos x}} = \frac{(\sqrt{1 + cos x} + \sqrt{1 - cos x})^{2}}{(\sqrt{1 + cos x})^{2} - (\sqrt{1 - cos x})^{2}}$

$= \frac{1 + cos x + 1 - cos x + 2 \sqrt{1 + cos x} \sqrt{1 - cos x}}{(1 + cos x) - (1 - cos x)}$

$= \frac{2 + 2 \sqrt{1 - {cos}^{2} x}}{2 cos x}$

$= \frac{2 + 2 sin x}{2 cos x}$

$= \frac{2}{2 cos x} + \frac{2 sin x}{2 cos x}$

$= sec x + tan x$

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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

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})

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b

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 is

√1+cosx+√1−cosx√1+cosx−√1−cosx=(√1+cosx+√1−cosx)2(√1+cosx)2−(√1−cosx)2=1+cosx+1−cosx+2√1+cosx√1−cosx(1+cosx)−(1−cosx)=2+2√1−cos2x2cosx=2+2sinx2cosx=22cosx+2sinx2cosx=secx+tanx

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