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BITSAT 2017 Slot 2 Solved Paper

Section: Mathematics

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Question : 113 of 150

Marks: +1, -0

If

\cos (x-y), \cos x

\cos (x+y)

are in HP, then

\cos x \sec \left(\frac{y}{2}\right)

is equal to

$\pm \sqrt{2}$
$\pm \frac{1}{\sqrt{2}}$
$\pm 2$
None of these

Solution:

Given that,

\cos (x-y), \cos x, \cos (x+y)

are in HP.

Then,

\cos x=\frac{2 \cos (x-y) \cos (x+y)}{\cos (x-y)+\cos (x+y)}

\Rightarrow \cos x=\frac{2(\cos^2 x-\sin^2 y)}{2 \cos x \cos y}

\Rightarrow \cos^2 x \cos y=\cos^2 x-\sin^2 y

\Rightarrow \cos^2 x (\cos y-1)=-\sin^2 y

\Rightarrow \cos^2 x (1-\cos y)=1-\cos^2 y

\Rightarrow \cos^2 x (1-\cos y)=(1-\cos y)

(1+\cos y)

\Rightarrow \cos^2 x=1+\cos y \Rightarrow \cos^2 x

=\frac{2 \cos^2 y}{2}

\Rightarrow \cos^2 x \sec^2 \left(\frac{y}{2}\right)=2

\Rightarrow \cos x \sec \left(\frac{y}{2}\right)=\pm \sqrt{2}

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