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AP DSC School Assistant Math Model Test 1

Section: Subject Specific

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Question : 111 of 160

Marks: +1, -0

What is value of cot^{2} 15° + tan^{2} 15° ?

12
14
$8\sqrt{3}$
4

Solution:

Concept:

We use the standard trigonometric values:

\tan 15^\circ = 2 - \sqrt{3}

and

\cot 15^\circ = 2 + \sqrt{3}

, along with the algebraic identities

(a+b)^2 = a^2 + 2ab + b^2

and

(a-b)^2 = a^2 - 2ab + b^2

.

Explanation:

Let

y = \cot^2 15^\circ + \tan^2 15^\circ

. Substituting the known values:

y = (2 + \sqrt{3})^2 + (2 - \sqrt{3})^2

Now expand each square:

(2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}

(2 - \sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3}

Add them together:

y = (7 + 4\sqrt{3}) + (7 - 4\sqrt{3}) = 14

The

4\sqrt{3}

terms cancel, leaving

14

.

Answer:

Option B: 14

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