1)Find the maximum and minimum value of (1+sinA)(1+cosA)....
2)If tanA=4tanB, Find the maximum value of tan^2 (A-B)
Note - both are different questions.....!

I'll take A=x and B=y as I like x and y more than A and B.
\[1)f(x)=(1+sin x)(1+cos x)\]
\[f'(x)=(cos x-sin x)(1+cos x+sin x)\]
For stationary points:-
\[f'(x)=0;\]
\(x=\frac{\pi}{4}\) or \(x=\frac{-\pi}{2}\)

For maximum f"(x)<0 and for minimum f"(x)>0.
* So max.* \(f(x)=1.5+\sqrt2\) and min. \(f(x)=0\)
\[2)x=tan^{-1}(4tan y)\]
\[f(y)=tan^2(tan^{-1}(4tan y)-y)\]
\[f'(y)=2tan(x-y)sec^2(x-y)(\frac{4sec^2y}{1+16tan^2y} -(-1) )\]
For f'(y)=0
\[tan^2y=\frac{1}{4}\]
\(tan y=\frac{1}{2}\) and \(tan x=2\)
\[tan^2(x-y)=(\frac{tan x- tan y}{1+ tan x tan y})^2\]
Substituting maximum \(tan^2(x-y)= \frac{9}{16}\)

You working is wrong. You should be very careful when solving trigonometric equations. Make sure you consider all possible solutions. For example, you missed out several cases for 1. E.g. the minimum value of 0 can occur when \(\cos x = -1\) or when \(\sin x = -1 \), which shows that you have missed some cases. A similar situation holds for the maximum.

Similarly, while it is given that \( \tan x = 4 \tan y \), this does not imply that \( x = \tan^{-1} ( 4 \tan y) \), as this assumes that \(x\) lies in the principle domain. You will need to check other cases, and it seems that you simply got lucky. The answer could be very different if the condition was \( \tan 3x = \tan 4 y \), which need not imply that \(3x = 4y \).

In such type of questions usually the domain is given in question itself. As no domain was given, so I assumed principle domain to be the domain being considered in the question.

@Shubham Srivastava
–
Since no domain is stated, the implicit assumption is all values where the function is defined, which is all real numbers. I'm ignoring that trigonometric functions can be extended to the complex numbers, since they will be complex valued and won't have a maximum and minimum.

I am concerned about your logical argument, and highlighting an error in your logic. Note further that the principle domain of each of the trig functions are different, and hence you are merely looking at their intersection. Your answer is thus applicable to an (arbitrary) choice of domain which is \( [0, \frac{\pi}{2} ) \). This limits its usefulness, since in part the domain excludes the minimum that you calculated.

Related: What are the maximum and minimum values of \( x + \sin x \)?

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## Comments

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TopNewestI'll take A=x and B=y as I like x and y more than A and B. \[1)f(x)=(1+sin x)(1+cos x)\] \[f'(x)=(cos x-sin x)(1+cos x+sin x)\] For stationary points:- \[f'(x)=0;\] \(x=\frac{\pi}{4}\) or \(x=\frac{-\pi}{2}\)

For maximum f"(x)<0 and for minimum f"(x)>0.

* So max.* \(f(x)=1.5+\sqrt2\)and min.\(f(x)=0\) \[2)x=tan^{-1}(4tan y)\] \[f(y)=tan^2(tan^{-1}(4tan y)-y)\] \[f'(y)=2tan(x-y)sec^2(x-y)(\frac{4sec^2y}{1+16tan^2y} -(-1) )\]For f'(y)=0 \[tan^2y=\frac{1}{4}\] \(tan y=\frac{1}{2}\) and \(tan x=2\) \[tan^2(x-y)=(\frac{tan x- tan y}{1+ tan x tan y})^2\] Substituting

maximum\(tan^2(x-y)= \frac{9}{16}\)Log in to reply

You working is wrong. You should be very careful when solving trigonometric equations. Make sure you consider all possible solutions. For example, you missed out several cases for 1. E.g. the minimum value of 0 can occur when \(\cos x = -1\) or when \(\sin x = -1 \), which shows that you have missed some cases. A similar situation holds for the maximum.

Similarly, while it is given that \( \tan x = 4 \tan y \), this does not imply that \( x = \tan^{-1} ( 4 \tan y) \), as this assumes that \(x\) lies in the principle domain. You will need to check other cases, and it seems that you simply got lucky. The answer could be very different if the condition was \( \tan 3x = \tan 4 y \), which need not imply that \(3x = 4y \).

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In such type of questions usually the domain is given in question itself. As no domain was given, so I assumed principle domain to be the domain being considered in the question.

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I am concerned about your logical argument, and highlighting an error in your logic. Note further that the principle domain of each of the trig functions are different, and hence you are merely looking at their intersection. Your answer is thus applicable to an (arbitrary) choice of domain which is \( [0, \frac{\pi}{2} ) \). This limits its usefulness, since in part the domain excludes the minimum that you calculated.

Related: What are the maximum and minimum values of \( x + \sin x \)?

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thanxxxx........

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if (1-3x)^1/2+(1-x)^5/3 (whole divided by) (4-x)........ is approximately equal to a+bx.find a and b.

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