MHTCET 25 Apr 2025 Shift 1 Paper

Question : 102 of 150

Marks: +1, -0

\sec x+\tan x=2,0 < x < \frac{\pi}{2}

\sin \frac{x}{4}=

Solution:

Concept:

Use the identity

\sec^2 x - \tan^2 x = 1

to find

\cos x

, then apply half-angle formulas twice to get

\sin \frac{x}{4}

Explanation:

Given

\sec x+\tan x=2

Using

\sec^2 x-\tan^2 x=1

(\sec x+\tan x)(\sec x-\tan x)=1

\sec x-\tan x=\frac{1}{2}

Adding both equations:

2\sec x=\frac{5}{2}

\sec x=\frac{5}{4}

, hence

\cos x=\frac{4}{5}

Now

0 < x < \frac{\pi}{2}

, so

\frac{x}{2}

lies in the first quadrant.

Thus,

\sin\frac{x}{2}

and

\cos\frac{x}{2}

are positive.

\sin\frac{x}{2}=\sqrt{\frac{1-\cos x}{2}}

=\sqrt{\frac{1-\frac{4}{5}}{2}}=\frac{1}{\sqrt{10}}

\cos^2\frac{x}{2}=1-\sin^2\frac{x}{2}

=1-\frac{1}{10}=\frac{9}{10}

\cos\frac{x}{2}=\frac{3}{\sqrt{10}}

Finally,

\sin\frac{x}{4}=\sqrt{\frac{1-\cos\frac{x}{2}}{2}}

=\sqrt{\frac{1-\frac{3}{\sqrt{10}}}{2}}

=\sqrt{\frac{10-3\sqrt{10}}{20}}

=\frac{1}{\sqrt{2(10+3\sqrt{10})}}

Answer:

\sin\frac{x}{4}=\frac{1}{\sqrt{2(10+3\sqrt{10})}}

, which is option B.

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