Question

Evaluate the following: $\dfrac{\tan 45{}^\circ }{\cos \text{ec}30{}^\circ }+\dfrac{\sec 60{}^\circ }{\cot 45{}^\circ }-\dfrac{5\sin 90{}^\circ }{2\cos 0{}^\circ }$ .

Answer 1

Hint: Use Trigonometric ratios table to find the values of trigonometric standard angles such as $0{}^\circ ,30{}^\circ ,45{}^\circ ,60{}^\circ \text{ and 90}{}^\circ $ . It consists of trigonometric ratios – sine, cosine, tangent, cosecant, secant and cotangent. These ratios can be written in short as sin, cos, tan, cosec, sec and cot. So, just put the values of the required trigonometric ratios in the expression given and solve to get the answer.

Complete step by step solution:
Before moving to the solution, let us discuss the nature of sine and cosine function, which we would be using in the solution. We can better understand this using the graph of sine and cosine.
First, let us start with the graph of sinx.

Next, let us see the graph of cosx.

Looking at both the graphs, and using the relations between the different trigonometric ratios, we get

Now to start with the solution to the above question, we will try to simplify the expression given in the question by putting the values $\cos 45{}^\circ =\dfrac{1}{\sqrt{2}}$ , $\cot 45{}^\circ =1$ and $\tan 45{}^\circ =1$ . On doing so, we get
$\dfrac{\tan 45{}^\circ }{\cos \text{ec}30{}^\circ }+\dfrac{\sec 60{}^\circ }{\cot 45{}^\circ }-\dfrac{5\sin 90{}^\circ }{2\cos 0{}^\circ }$
$=\dfrac{1}{\cos \text{ec}30{}^\circ }+\dfrac{\sec 60{}^\circ }{1}-\dfrac{5\sin 90{}^\circ }{2\cos 0{}^\circ }$
Now we also know that $\sec 60{}^\circ =2$ , \[\text{cosec30 }\!\!{}^\circ\!\!\text{ }=2\] and $\cos 0{}^\circ =1$ . So, if we put these values in our expression, we get
$=\dfrac{1}{2}+\dfrac{2}{1}-\dfrac{5\sin 90{}^\circ }{2}$
Finally, we will put $\sin 90{}^\circ =1$ . On putting the values in our expression, we get
$=\dfrac{1}{2}+\dfrac{2}{1}-\dfrac{5}{2}$
$=0$
Therefore, we can say that the value of $\dfrac{\tan 45{}^\circ }{\cos \text{ec}30{}^\circ }+\dfrac{\sec 60{}^\circ }{\cot 45{}^\circ }-\dfrac{5\sin 90{}^\circ }{2\cos 0{}^\circ }$ is equal to $0$ .

Note: The most important thing that you need to remember is the values of different trigonometric ratios for different standard angles. Remembering the graphs of different trigonometric ratios may also be helpful in some of the questions. Also, while putting the values of different trigonometric ratios in the expression, don’t get confused between the values of different trigonometric ratios.