If -1-cos x-1-cos x-1-cos x-1-cos x-a-x-2- xe- 2 - then a

The correct option is A π/4 √(1+cos x)=√(2 cos^2 x/2)=√(2)|cos x/2| √(1 cos x)=√(2 sin^2 x/2)=√(2)|sin x/2| π lt; x lt; 2π⇒π/2 lt; x/2 lt; π. ∴√(1+cos ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Trigonometric Ratios of Multiples of an Angle

If √1+cos x+√...

Question

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___

$\frac{π}{4}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$\frac{π}{3}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{π}{2}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\frac{2 π}{3}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
$\frac{π}{4}$

$\sqrt{1 + c o s x} = \sqrt{2 c o s^{2} \frac{x}{2}} = \sqrt{2} | c o s \frac{x}{2} | \sqrt{1 - c o s x} = \sqrt{2 s i n^{2} \frac{x}{2}} = \sqrt{2} | s i n \frac{x}{2} | π < x < 2 π \Rightarrow \frac{π}{2} < \frac{x}{2} < π . ∴ \sqrt{1 + c o s x} = - c o s \frac{x}{2} \sqrt{1 - c o s x} = s i n \frac{x}{2}$
$∴ \frac{\sqrt{1 + c o s x} + \sqrt{1 - c o s x}}{\sqrt{1 + c o s x} - \sqrt{1 - c o s x}} = \frac{- c o s \frac{x}{2} + s i n \frac{x}{2}}{- c o s \frac{x}{2} - s i n \frac{π}{2}} \Rightarrow \frac{- 1 t a n \frac{x}{2}}{- 1 - t a n \frac{x}{2}} = \frac{1 - t a n \frac{x}{2}}{1 + t a n \frac{x}{2}} = t a n (\frac{π}{4} - \frac{x}{2}) = c o t (\frac{π}{2} - (\frac{π}{4} - \frac{x}{2})) = c o t (\frac{π}{4} + \frac{x}{2})$

Suggest Corrections

Similar questions

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. If

π < x < 2 π

then

\frac{\sqrt{1 + cos x} + \sqrt{1 - cos x}}{\sqrt{1 + cos x} - \sqrt{1 - cos x}} =

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 cos $\frac{x}{2}$ , cos $\frac{x}{2^{2}}$ , cos $\frac{x}{2^{3}}$ .............cos $\frac{x}{2^{n}}$ = $\frac{s i n x}{2^{n} s i n \frac{x}{2^{n}}}$ Then

$\frac{1}{2}$ tan $\frac{x}{2}$ + $\frac{1}{2^{2}}$ tan $\frac{x}{2^{2}}$ + ............. $\frac{1}{2^{n}}$ tan $\frac{x}{2^{n}}$ is

Join BYJU'S Learning Program

If √1+cosx+√1−cosx√1+cosx=√1−cosx=cot(a+x2),xϵ(π,2π) then a___

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___