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The minimum value of |sinx+cosx+tanx+secx+cosecx+cotx| is
A) 221
B) 22+1
C) 21
D) 2+1

Answer
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Hint:
Here, we will first convert all these functions in terms of the sine and cosine. Then we will add them together and simplify it further to find the minimum value of the sum. Trigonometric ratios are defined as the ratios of any two sides of a right angled triangle.

Formula Used:
We will use the following Formula:
1) Trigonometric ratio:tanx=sinxcosx;secx=1cosx;cosecx=1sinx;cotx=cosxsinx
2) Trigonometric identities: sin2x+cos2x=1
3) The square of the sum of two numbers is given by the formula: (a+b)2=a2+b2+2ab
4) The difference of square of two numbers is given by the algebraic identity (a2b2)=(a+b)(ab)
5) Trigonometric formula: sin(A+B)=sinAcosB+cosAsinB
6) |x|={x,ifx0x,ifx<0

Complete step by step solution:
We are given with a trigonometric function |sinx+cosx+tanx+secx+cosecx+cotx|
Let |sinx+cosx+tanx+secx+cosecx+cotx|=y
Now, we have to represent all the terms in terms of sine and cosine.
Now, substituting tanx=sinxcosx,secx=1cosx,cosecx=1sinx and cotx=cosxsinx in the given expression, we get
y=sinx+cosx+sinxcosx+1cosx+1sinx+cosxsinx

By grouping the terms and taking L.C.M., we get
y=sinx+cosx+sinxcosx+cosxsinx+1sinx+1cosx
y=sinx+cosx+sinxcosx×sinxsinx+cosxsinx×cosxcosx+1sinx×cosxcosx+1cosx×sinxsinx
Adding the like terms, we get
y=sinx+cosx+sin2x+cos2xcosxsinx+cosx+sinxcosxsinx
Now, by using the trigonometric identity sin2x+cos2x=1, we get
 y=sinx+cosx+1+cosx+sinxcosxsinx ………………………..(1)
Using the algebraic identity (a+b)2=a2+b2+2ab, we can write
(sinx+cosx)2=sin2x+cos2x+2sinxcosx
(sinx+cosx)2=1+2sinxcosx
Rewriting the equation, we get
sinxcosx=(sinx+cosx)212
Substituting sinxcosx=(sinx+cosx)212 in equation (1), we get
y=sinx+cosx+1+cosx+sinx(sinx+cosx)212
Substituting sinx+cosx=t in the above equation, we get
sinx+cosx+1+cosx+sinx(sinx+cosx)212=t+1+tt212
sinx+cosx+1+cosx+sinx(sinx+cosx)212=t+2(1+t)t21
The difference of square of two numbers is given by the algebraic identity (a2b2)=(a+b)(ab)
sinx+cosx+1+cosx+sinx(sinx+cosx)212=t+2(1+t)(1+t)(t1)
sinx+cosx+1+cosx+sinx(sinx+cosx)212=t+2(t1)………………………..(2)
We know that t=sinx+cosx.
Multiplying and dividing by 2on both the sides, we get
t=22sinx+22cosx=2(12sinx+12cosx)
t=2(cosπ4sinx+sinπ4cosx)
Now, by using trigonometric formula sin(A+B)=sinAcosB+cosAsinB, we get
t=2(sin(x+π4))
Thus , by this identity, we will get 2t2 .
Substituting y=|sinx+cosx+tanx+secx+cosecx+cotx|=t+2(t1)
y=|t+2t1|=|t1+2t1+1|
Now, by using the property |x|={x,ifx0x,ifx<0 , we get
Now, considering 1<a<20<a1in this case the value is positive.
y=t1+2t1+1
We know that Arithmetic mean Geometric mean. Thus, we will get
t1+2t12(t1)×2t1
t1+2t122t1+2t122
t1+2t1+122+1
Thus, we get
y22+1 which is positive…………………………………………………..(3)
Now, considering 1<a<2a1<0 in this case the value is negative
y=t1+2t1+1
y=|1(1t+21t1)|
y=|1||(1t+21t1)|
y=|(1t+21t1)|
We know that Arithmetic mean Geometric mean. Thus, we will get
1t+21t2(1t)×21t
1t+21t221t+21t22
1t+21t1221
Thus, we get
y221 which is positive………………………………………………(4)
Since the value has to be minimum, from (3) and (4), we get 221<22+1
Thus the minimum value is 221

Therefore, the minimum value of |sinx+cosx+tanx+secx+cosecx+cotx| is 221.

Note:
We can find the limits by using the trigonometric identity and values. So, it becomes important to remember all the basic identities and values. We have found the minimum value using the relation between arithmetic mean and geometric mean. Arithmetic mean or average is defined as the sum of numbers divided by the quantity of numbers, Geometric mean of two numbers is defined as the square root of their product. We should also remember that the Arithmetic mean should never be less than the geometric mean.
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