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Telangana TET Paper 2 Math and Science Model Paper 2

Section: Math and Science

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Question : 131 of 150

Marks: +1, -0

If

\frac{sec\alpha +tan\alpha }{sec\alpha -tan\alpha }=\frac{7}{4}

, then find the value of cosec α.

$\frac{1}{3}$
$\frac{1}{11}$
$\frac{11}{3}$
$\frac{3}{11}$

Solution:

Concept:

Use the Componendo and Dividendo rule: if

a/b = c/d

, then

(a+b)/(a-b) = (c+d)/(c-d)

. Also recall

\sec\alpha = 1/\cos\alpha

,

\tan\alpha = \sin\alpha/\cos\alpha

, and

\csc\alpha = 1/\sin\alpha

.

Explanation:

Given

\frac{\sec\alpha + \tan\alpha}{\sec\alpha - \tan\alpha} = \frac{7}{4}

.

Apply Componendo and Dividendo:

\frac{(\sec\alpha+\tan\alpha)+(\sec\alpha-\tan\alpha)}{(\sec\alpha+\tan\alpha)-(\sec\alpha-\tan\alpha)} = \frac{7+4}{7-4}

.

Simplify numerator to

2\sec\alpha

and denominator to

2\tan\alpha

. So

\frac{2\sec\alpha}{2\tan\alpha} = \frac{11}{3}

, i.e.,

\frac{\sec\alpha}{\tan\alpha} = \frac{11}{3}

.

Express in terms of sine:

\frac{\sec\alpha}{\tan\alpha} = \frac{1/\cos\alpha}{\sin\alpha/\cos\alpha} = \frac{1}{\sin\alpha} = \csc\alpha

.

Thus

\csc\alpha = \frac{11}{3}

.

Alternatively, cross-multiply:

4(\sec\alpha+\tan\alpha)=7(\sec\alpha-\tan\alpha)

→

11\tan\alpha=3\sec\alpha

→

11\frac{\sin\alpha}{\cos\alpha}=3\frac{1}{\cos\alpha}

→

11\sin\alpha=3

→

\sin\alpha=3/11

→

\csc\alpha=1/\sin\alpha=11/3

.

Answer:

Option C:

\frac{11}{3}

.

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