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AP and TS EAMCET Engineering Model Paper 1

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Question : 14 of 160

Marks: +1, -0

In a triangle

A B C,

\tan A : \tan B : \tan C=2 : 3 : 4

\sec^{2} C=

17
7
9
11

Solution:

For a triangle,

A+B+C=\pi

, so the tangent identity gives

\tan A+\tan B+\tan C=\tan A\tan B\tan C

.

Let

\tan A=2k

,

\tan B=3k

, and

\tan C=4k

.

Then

\tan A+\tan B+\tan C=9k

.

Also

\tan A\tan B\tan C=(2k)(3k)(4k)=24k^3

.

Thus

9k=24k^3

.

Since

k\neq 0

,

k^2=\frac{3}{8}

.

Therefore

\tan^2 C=(4k)^2=16k^2=16\cdot\frac{3}{8}=6

.

Using

\sec^2 C=1+\tan^2 C

, we get

\sec^2 C=1+6=7

.

Hence the correct option is

7

.

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