Question
Question: Using tables evaluate the following: \(4\cot 45^\circ - {\sec ^2}60^\circ + \sin 30^\circ \)...
Using tables evaluate the following:
4cot45∘−sec260∘+sin30∘
Solution
Start by writing the equation. Try to recall all the values of trigonometric ratios for corresponding angles and substituting in the given equation. After that we simplify the given equation in order to get the required result for the given expression.
Complete step-by-step solution:
It is given that the question stated as, 4cot45∘−sec260∘+sin30∘
Here we have to find the value for the equation
4cot45∘−sec260∘+sin30∘
We can use the trigonometric table to solve the above expression. The trigonometric table helps us to find the values of the trigonometric ratios sine, cosine, tangent, cosecant, secant and cotangent that is they can be written as sin, cos, tan, cosec, sec, and cot.
To find the value for the above trigonometric expression,
We need to know the trigonometric values for the different angles.
The values of the trigonometric ratios as follows:
Angles(degree) | 0∘ | 30∘ | 45∘ | 60∘ | 90∘ |
---|---|---|---|---|---|
Angles(radian) | 0 | 6π | 4π | 3π | 2π |
sinθ | 0 | 21 | 21 | 23 | 1 |
cosθ | 1 | 23 | 21 | 21 | 0 |
tanθ | 0 $$ | 31 | 1 | 3 | Not defined |
cotθ | Not defined | 3 | 1 | 31 | 0 |
secθ | 1 | 32 | 2 | 2 | Not defined |
cosecθ | Not defined | 2 | 2 | 32 | 1 |
From the table we can write it as,
cot45∘=1 , sin60∘=2 and sin30∘=21
Now we have the given equation, 4cot45∘−sec260∘+sin30∘
Substituting these values in the relation as below:
⇒4×1−(2)2+21
By doing some elementary operations like multiply and square of the number, 2
Hence we get,
⇒4−4+21
Now, we can cancel the opposite numbers, and we get
⇒21
So the value of 4cot45∘−sec260∘+sin30∘ is 21.
Hence the correct answer is 21
Note: Relation between different trigonometric functions is as follows:
sinA=cosecA1
cosA=secA1
secA=cosA1
cosecA=sinA1
Students must remember all the trigonometric values of different angles, which are used more often. Attention must be given while substituting the values, keeping in mind the quadrant system, here we had all the positive values only, but one might get different quadrant values as well.