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UPSC CDS 1 2022 Math Paper

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Question : 5 of 100

Marks: +1, -0

What is the value of the following

\frac{2 \sin 68^{\circ}}{\cos 22^{\circ}} - \frac{2 \cot 15^{\circ}}{5 \tan 75^{\circ}} - \frac{3 \tan 20^{\circ} \tan 40^{\circ} \tan 45^{\circ} \tan 50^{\circ} \tan 70^{\circ}}{5}

$-1$
0
1
5

Solution: 👈: Video Solution

Concept:

Using trigonometric ratios of complementary angles:-

-

\sin (90^{\circ} - heta) = \cos heta

-

\cos (90^{\circ} - heta) = \sin heta

-

an (90^{\circ} - heta) = \cot heta

-

\cot (90^{\circ} - heta) = an heta

-

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

Calculation:

\frac{2 \sin 68^{\circ}}{\cos 22^{\circ}} - \frac{2 \cot 15^{\circ}}{5 \tan 75^{\circ}} - \frac{3 \tan 20^{\circ} \tan 40^{\circ} \tan 45^{\circ} \tan 50^{\circ} \tan 70^{\circ}}{5}

\Rightarrow \frac{2 \sin 68^{\circ}}{\cos (90^{\circ} - 68^{\circ})} - \frac{2 \cot 15^{\circ}}{5 \tan (90^{\circ} - 15^{\circ})} - \frac{3 \tan (90^{\circ} - 70^{\circ}) \tan (90^{\circ} - 50^{\circ}) \tan 45^{\circ} \tan 50^{\circ} \tan 70^{\circ}}{5}

\Rightarrow \frac{2 \sin 68^{\circ}}{\sin 68^{\circ}} - \frac{2 \cot 15^{\circ}}{5 \cot 15^{\circ}} - \frac{3 \tan 45^{\circ} \cot 70^{\circ} \cot 50^{\circ} \tan 50^{\circ} \tan 70^{\circ}}{5}

\Rightarrow \frac{2 \sin 68^{\circ}}{\sin 68^{\circ}} - \frac{2 \cot 15^{\circ}}{5 \cot 15^{\circ}} - \frac{3 \tan 45^{\circ} \cdot \frac{1}{\tan 70^{\circ}} \cdot \frac{1}{\tan 50^{\circ}} \cdot \tan 50^{\circ} \cdot \tan 70^{\circ}}{5}

\Rightarrow \frac{2 \sin 68^{\circ}}{\sin 68^{\circ}} - \frac{2 \cot 15^{\circ}}{5 \cot 15^{\circ}} - \frac{3 \tan 45^{\circ}}{5}

\Rightarrow 2 - \frac{2}{5} - \frac{3 \times 1}{5} \qquad [\because \tan 45^{\circ}=1]

\Rightarrow 2 - \frac{5}{5}

\Rightarrow 2 - 1 = 1

herefore

The value of

\frac{2 \sin 68^{\circ}}{\cos 22^{\circ}} - \frac{2 \cot 15^{\circ}}{5 \tan 75^{\circ}} - \frac{3 \tan 20^{\circ} \tan 40^{\circ} \tan 45^{\circ} \tan 50^{\circ} \tan 70^{\circ}}{5}

is

1

.

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