Question

Without using trigonometric tables, evaluate the following:
\[2\left( {\dfrac{{\cos {{58}^0}}}{{\sin {{32}^0}}}} \right) - \sqrt 3 \left( {\dfrac{{\cos {{38}^0}\cos ec{{52}^0}}}{{\tan {{15}^0}\tan {{60}^0}\tan {{75}^0}}}} \right).\]

Answer 1

Hint: To attempt this question remember the trigonometric identities and remember to use $\tan ({90^0} - \theta ) = \cot \theta $ and $\cos ({90^0} - \theta ) = \sin \theta $in the equation then apply the identities like $\sin \theta = \dfrac{1}{{\cos ec\theta }}$and $\tan \theta = \dfrac{1}{{\cot \theta }}$, use this information to approach the solution.

Complete step-by-step answer:
According to the given information we have the function \[2\left( {\dfrac{{\cos {{58}^0}}}{{\sin {{32}^0}}}} \right) - \sqrt 3 \left( {\dfrac{{\cos {{38}^0}\cos ec{{52}^0}}}{{\tan {{15}^0}\tan {{60}^0}\tan {{75}^0}}}} \right).\]
Let us assume that
$I = 2\left( {\dfrac{{\cos {{58}^0}}}{{\sin {{32}^0}}}} \right) - \sqrt 3 \left( {\dfrac{{\cos {{38}^0}\cos ec{{52}^0}}}{{\tan {{15}^0}\tan {{60}^0}\tan {{75}^0}}}} \right)$ (equation 1)
Since we know that $\tan ({90^0} - \theta ) = \cot \theta $ and $\cos ({90^0} - \theta ) = \sin \theta $
Using this in the above equation we get
$2\left( {\dfrac{{\cos ({{90}^0} - {{32}^0})}}{{\sin {{32}^0}}}} \right) - \sqrt 3 \left( {\dfrac{{\cos ({{90}^0} - {{52}^0})\cos ec{{52}^0}}}{{\tan {{15}^0}\tan {{60}^0}\tan ({{90}^0} - {{15}^0})}}} \right)$
$ \Rightarrow $$2\left( {\dfrac{{\sin {{32}^0})}}{{\sin {{32}^0}}}} \right) - \sqrt 3 \left( {\dfrac{{\sin {{52}^0}\cos ec{{52}^0}}}{{\tan {{15}^0} \times \sqrt 3 \times \cot {{15}^0}}}} \right)$
Now, we know that $\sin \theta = \dfrac{1}{{\cos ec\theta }}$ or $\sin \theta \cos ec\theta = 1$
We also know that,$\tan \theta = \dfrac{1}{{\cot \theta }}$ or $\tan \theta \cot \theta = 1$
Using these identities in the equation 1 we get
$I = 2(1) - \sqrt 3 \left( {\dfrac{1}{{\sqrt 3 }}} \right)$
$ \Rightarrow $$I = 2 - 1 = 1$
Therefore, the value of given function i.e. $2\left( {\dfrac{{\cos {{58}^0}}}{{\sin {{32}^0}}}} \right) - \sqrt 3 \left( {\dfrac{{\cos {{38}^0}\cos ec{{52}^0}}}{{\tan {{15}^0}\tan {{60}^0}\tan {{75}^0}}}} \right) = 1$
So, this is the required answer.

Note: In the above solution we used the trigonometric identities which are the expressions which involve trigonometric functions where the term “function” can be explained as relation between the provided inputs and the outputs of the given inputs such that each input is directly related to the one output. The representation of a function is given by supposing if there is a function “f” that belongs from X to Y then the function is represented by $f:X \to Y$ examples of function are one-one functions, onto functions, bijective functions, trigonometric function, binary function, etc.