Trigonmetry is fun

Geometry Level 2

If A + B = π 4 A+B = \frac{\pi}4 , find the value of ( 1 + tan ⁡ A ) ( 1 + tan ⁡ B ) (1 +\tan A )(1 +\tan B ) .

Note : A , B ≠ π 2 + π n A,B\ne \frac\pi2 + \pi n for integer n n .


The answer is 2.

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Moinul Islam Tanvir by Moinul Islam Tanvir , 23, Bangladesh

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1 solution

Nihar Mahajan
Sep 26, 2015

tan ⁡ ( A + B ) = tan ⁡ A + tan ⁡ B 1 − tan ⁡ A tan ⁡ B ⇒ tan ⁡ ( π 4 ) = tan ⁡ A + tan ⁡ B 1 − tan ⁡ A tan ⁡ B ⇒ 1 = tan ⁡ A + tan ⁡ B 1 − tan ⁡ A tan ⁡ B ⇒ 1 − tan ⁡ A tan ⁡ B = tan ⁡ A + tan ⁡ B ⇒ 1 = tan ⁡ A + tan ⁡ B + tan ⁡ A tan ⁡ B ⇒ 2 = tan ⁡ A + tan ⁡ B + tan ⁡ A tan ⁡ B + 1 ⇒ 2 = 1 ( 1 + tan ⁡ A ) + tan ⁡ B ( 1 + tan ⁡ A ) ⇒ 2 = ( 1 + tan ⁡ A ) ( 1 + tan ⁡ B ) \Large{\begin{aligned} \tan(A+B) &= \dfrac{\tan A + \tan B}{1-\tan A\tan B} \\ \Rightarrow \tan\left(\dfrac{\pi}{4}\right) &= \dfrac{\tan A + \tan B}{1-\tan A\tan B} \\ \Rightarrow 1 &= \dfrac{\tan A + \tan B}{1-\tan A\tan B} \\ \Rightarrow 1-\tan A\tan B &=\tan A + \tan B \\ \Rightarrow 1 &= \tan A + \tan B+\tan A\tan B \\ \Rightarrow 2 &= \tan A + \tan B+\tan A\tan B+1\\ \Rightarrow 2 &=1(1+\tan A) + \tan B(1+\tan A) \\ \Rightarrow \boxed{2} &=(1+\tan A )(1+\tan B) \end{aligned} }

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Shortcut :
Given relation is an identity, it means it must be true for all real numbers in its domain.

⇒ A = 0 , B = π 4 \Rightarrow A = 0 \ , \ B = \dfrac{\pi}{4}
⇒ ( 1 + tan ⁡ 0 ) ( 1 + tan ⁡ π 4 ) \Rightarrow (1 + \tan 0)(1 + \tan\dfrac{\pi}{4})
⇒ ( 1 + 0 ) ( 1 + 1 ) \Rightarrow (1 + 0)(1 + 1)
⇒ 2 \Rightarrow \boxed{\color{#3D99F6}2}

Akhil Bansal - 5 years, 8 months ago

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This solution is incomplete. It does not show that the expression equals 2 for all possible values of A , B A,B .

Nihar Mahajan - 5 years, 8 months ago

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If domain is not given in the equation, it means given function must be true for all values of x in its own domain.

Akhil Bansal - 5 years, 8 months ago

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@Akhil Bansal – Your way of doing this is suited for examinations like JEE where you have to do a lot of problems in a short amount of time. But, mathematically speaking, your solution is indeed incomplete.

You're assuming that the answer to this problem is unique to solve this (which is always the case for a problem on Brilliant but not always for a problem in general).

Let's say you're given the following problem :

A + B = π 2 tan ⁡ ( A ) + tan ⁡ ( B ) = ? ? ? A+B=\frac{\pi}{2}\qquad \tan(A)+\tan(B)=???

Can you use the same trick here? No. Why? Because first of all, there are multiple answers to the problem and secondly, there are singularities at points like ( A , B ) ∈ { ( 0 , π 2 ) , ( π 2 , 0 ) } (A,B)\in\left\{\left(0,\frac{\pi}2\right),\left(\frac{\pi}2,0\right)\right\} .

So, to conclude, your method might be a shortcut but isn't a complete mathematical solution.

Prasun Biswas - 5 years, 8 months ago

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@Prasun Biswas – Yes you are correct,
I posted shortcut because actual solution is already been posted by Nihar and on brilliant, there are many students who are preparing for JEE (just like me) , So , they should know these type of tricks for quick problem solving.

Akhil Bansal - 5 years, 8 months ago

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