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16.
CBSE | Class 10 | Maths | RD Sharma | Trigonometric Ratios
16.

Solution:

Given, tan θ =

    \[1/\text{ }\surd 7\text{ }\ldots ..\left( 1 \right)\]

By definition, we know that

tan θ = Perpendicular side opposite to ∠θ / Base side adjacent to ∠θ

    \[\ldots \ldots \left( 2 \right)\]

On comparing equation

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

, we have

Perpendicular side opposite to ∠θ =

    \[1\]

Base side adjacent to ∠θ =

    \[\surd 7\]

Thus, the triangle representing ∠ θ is,

Hypotenuse AC is unknown and it can be found by using Pythagoras theorem

By applying Pythagoras theorem, we have

AC2 = AB2 + BC2

AC2 =

    \[{{1}^{2}}~+\text{ }{{\left( \surd 7 \right)}^{2}}\]

AC 2 =

    \[1\text{ }+\text{ }7\]

AC2 =

    \[8\]

AC =

    \[\surd 8\]

⇒ AC =

    \[2\surd 2\]

By definition,

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

⇒ sin θ =

    \[1/\text{ }2\surd 2\]

And, since cosec θ =

    \[1/\]

sin θ

⇒ cosec θ =

    \[2\surd 2\text{ }\ldots \ldots ..\text{ }\left( 3 \right)\]

Now,

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

⇒ cos θ =

    \[\surd 7/\text{ }2\surd 2\]

And, since sec θ =

    \[1/\]

sin θ

⇒ sec θ =

    \[2\surd 2/\text{ }\surd 7\text{ }\ldots \ldots .\text{ }\left( 4 \right)\]

Taking the L.H.S of the equation,

Substituting the value of cosec θ and sec θ from equation

    \[\left( 3 \right)\text{ }and\text{ }\left( 4 \right),\]

we get

i

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