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15.
CBSE | Class 10 | Maths | RD Sharma | Trigonometric Ratios
15.

Solution:

Given,

    \[cot\text{ }\theta \text{ }=\text{ }1/\surd 3\ldots \ldots .\text{ }\left( 1 \right)\]

By definition we know that,

    \[cot\text{ }\theta \text{ }=\text{ }1/\text{ }tan\text{ }\theta \]

And, since tan θ = perpendicular side opposite to ∠θ / Base side adjacent to ∠θ

cot θ = Base side adjacent to ∠θ / perpendicular side opposite to

    \[\angle \theta ~\ldots \ldots \text{ }\left( 2 \right)\]

[Since they are reciprocal to each other]On comparing equation

    \[\left( 1 \right)\text{ }and\text{ }\left( 2 \right),\]

we get

Base side adjacent to 

    \[\angle \theta ~=\text{ }1\]

and Perpendicular side opposite to 

    \[\angle \theta \text{ }=\text{ }\surd 3\]

Therefore, the triangle formed is,

On substituting the values of known sides as

    \[AB\text{ }=\text{ }\surd 3\text{ }and\text{ }BC\text{ }=\text{ }1\]

    \[A{{C}^{2~}}=\text{ }\left( \surd 3 \right)\text{ }+\text{ }1\]

    \[A{{C}^{2}}~=\text{ }3\text{ }+\text{ }1\]

    \[A{{C}^{2~}}=\text{ }4\]

    \[AC\text{ }=\text{ }\surd 4\]

Therefore,

    \[AC\text{ }=\text{ }2\text{ }\ldots \text{ }~~\left( 3 \right)\]

Now, by definition

sn θ = Perpendicular side opposite to ∠θ / Hypotenuse = AB / AC

    \[\Rightarrow sin\text{ }\theta \text{ }=\text{ }\surd 3/\text{ }2\text{ }\ldots \ldots \left( 4 \right)\]

And, cos θ = Base side adjacent to ∠θ / Hypotenuse = BC / AC

    \[\Rightarrowcos\text{ }\theta \text{ }=\text{ }1/\text{ }2\text{ }\ldots ..\text{ }\left( 5 \right)\]

Now, taking L.H.S we have

Substituting the values from equation

    \[\left( 4 \right)\text{ }and\text{ }\left( 5 \right),\]

we have

i

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