Test Index

UPSC NDA 2 2016 Mathematics Solved Paper

© examsnet.com

Question : 50 of 120

Marks: +1, -0

What is

\frac{{1 - \tan 2^\circ \cot 62^\circ }}{{\tan 152^\circ - \cot 88^\circ }}

equal to?

√3
-√3
$\sqrt{2} - 1$
$1 - \sqrt{2}$

Solution:

Concept:

Use complementary angle identities:

\tan(90^\circ - \theta) = \cot\theta

and

\tan(90^\circ + \theta) = -\cot\theta

. Also use the sum formula for cotangent:

\cot(x+y) = \frac{\cot x \cot y - 1}{\cot x + \cot y}

.

Explanation:

Start with the expression:

\frac{1 - \tan 2^\circ \cot 62^\circ}{\tan 152^\circ - \cot 88^\circ}

.

Rewrite

\tan 2^\circ

as

\cot(90^\circ - 2^\circ) = \cot 88^\circ

.

Rewrite

\tan 152^\circ

as

\tan(90^\circ + 62^\circ) = -\cot 62^\circ

.

The expression becomes:

\frac{1 - \cot 88^\circ \cot 62^\circ}{-\cot 62^\circ - \cot 88^\circ}

.

The denominator is

-(\cot 62^\circ + \cot 88^\circ)

.

So we have

\frac{1 - \cot 88^\circ \cot 62^\circ}{-(\cot 62^\circ + \cot 88^\circ)}

.

Multiply numerator and denominator by

-1

to get

\frac{\cot 88^\circ \cot 62^\circ - 1}{\cot 62^\circ + \cot 88^\circ}

.

This matches the formula

\cot(A+B) = \frac{\cot A \cot B - 1}{\cot A + \cot B}

with

A = 88^\circ

and

B = 62^\circ

.

Thus the expression equals

\cot(88^\circ + 62^\circ) = \cot 150^\circ

.

Now

\cot 150^\circ = \cot(90^\circ + 60^\circ) = -\tan 60^\circ = -\sqrt{3}

.

Answer:

-\sqrt{3}

(option B).

© examsnet.com

Go to Question:

Prev Question

Next Question