Question

Find the value of the expression $\dfrac{\sin {{53}^{0}}}{\cos {{37}^{0}}}+2\tan {{45}^{0}}-\dfrac{\cos ec{{60}^{0}}}{\sec {{30}^{0}}}$ .

Answer 1

Hint: Here, we will use certain trigonometric formulas, that is formulas for complementary angles to find the value of the given expression. The complementary angle formulas states that :

$\begin{align}

  & \sin \left( {{90}^{0}}-\theta \right)=\cos \theta \\

 & \cos ec\left( {{90}^{0}}-\theta \right)=\sec \theta \\

 & \tan \left( {{90}^{0}}-\theta \right)=\cot \theta \\

\end{align}$


Trigonometry is a branch of mathematics that studies the relationship between side lengths and angles of triangles. Trigonometric ratios are the ratios between edges of a right triangle. These ratios are given by trigonometric functions of the known angle A, where a, b and c refers to the lengths of the sides in the triangle given:-

Complete step-by-step answer:

Sine function is defined as the ratio of the perpendicular of the triangle to its hypotenuse.

$\sin A=\dfrac{perpendicular}{hypotenuse}=\dfrac{a}{c}$

Cosine function is defined as the ratio of the base of the triangle to its hypotenuse.

$\cos A=\dfrac{base}{hypotenuse}=\dfrac{b}{c}$

Tangent function is defined as the ratio of perpendicular to base of the triangle.

$\text{tan}A=\dfrac{perpendicular}{base}$

The hypotenuse is the side opposite to the 90 degree angle in a right angle triangle; it is the longest side of the triangle. The terms perpendicular and base are used for the opposite and adjacent sides respectively.

We also have:

$\begin{align}

  & \cos ecA=\dfrac{1}{\sin A} \\

 & \sec A=\dfrac{1}{\cos A} \\

 & \cot A=\dfrac{1}{\tan A} \\

\end{align}$

According to the complementary angle formula of trigonometry, we can write:
$\begin{align}

  & \sin \left( {{90}^{0}}-\theta \right)=\cos \theta \\

 & \cos ec\left( {{90}^{0}}-\theta \right)=\sec \theta \\

 & \tan \left( {{90}^{0}}-\theta \right)=\cot \theta \\

\end{align}$

Now, the expression given to us is:

$\dfrac{\sin {{53}^{0}}}{\cos {{37}^{0}}}+2\tan {{45}^{0}}-\dfrac{\cos ec{{60}^{0}}}{\sec {{30}^{0}}}$

Since, using complementary angle formulas we can write:

$\begin{align}

  & \cos {{37}^{0}}=\sin \left( {{90}^{0}}-{{37}^{0}} \right) \\

 & \Rightarrow \cos {{37}^{0}}=\sin {{53}^{0}} \\

\end{align}$

Also,

$\begin{align}

  & \sec {{30}^{0}}=\cos ec\left( {{90}^{0}}-{{30}^{0}} \right) \\

 & \Rightarrow \sec {{30}^{0}}=\cos ec{{60}^{0}} \\

\end{align}$

So, the given expression can be written as:

 $\begin{align}

  & \dfrac{\sin {{53}^{0}}}{\sin {{53}^{0}}}+2\tan {{45}^{0}}-\dfrac{\cos ec{{60}^{0}}}{\cos ec{{60}^{0}}} \\

 & =1+2\tan {{45}^{0}}-1 \\

\end{align}$

We know that the value of $\tan {{45}^{0}}=1$.

Therefore:

$1+2\tan {{45}^{0}}-1=1+2-1=2$

Hence, the value of the given expression is 2.


Note: Students should note here that we can calculate the value of the given expression by directly putting the values of the trigonometric ratios but the use of complementary angle formulas makes it easy to solve. Students should remember the values of standard trigonometric values like tan45 = 1.