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MHTCET 15 May 2024 Shift 2 Paper

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Question : 150 of 150

Marks: +1, -0

If

A+B=225^{\circ}

\frac{\cot A}{1+\cot A} \cdot \frac{\cot B}{1+\cot B}

, if it exists, is equal to

0
1
2
$\frac{1}{2}$

Solution:

Concept:

Use

\tan(A+B)

after converting

\cot

terms into

\tan

terms.

Explanation:

Given

A+B=225^\circ

, so

\tan(A+B)=\tan 225^\circ=1

.

Now rewrite each factor using

\cot A=\frac{1}{\tan A}

:

\frac{\cot A}{1+\cot A}=\frac{1}{1+\tan A}

and

\frac{\cot B}{1+\cot B}=\frac{1}{1+\tan B}

.

So the product becomes:

\frac{1}{(1+\tan A)(1+\tan B)}=\frac{1}{1+\tan A+\tan B+\tan A\tan B}

.

Since

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

,

we get

\tan A+\tan B=1-\tan A\tan B

.

Substitute this into the denominator:

1+\tan A+\tan B+\tan A\tan B=1+(1-\tan A\tan B)+\tan A\tan B=2

.

Hence the given expression equals

\frac{1}{2}

.

Answer:

\frac{1}{2}

Option D:

\frac{1}{2}

.

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