Fill in the blank. In a triangle ABC with =90^ the equation whose roots are tan A and tan B is _______. [Hint: A + B=90^ =1 and + =2 2A ]

In a triangle ABC with =90^ the equation whose roots are tan A and tan B is x^2- (2 2A )x+1=0. Solution: Given a with =90^ x^2-( + )x+ . =0 A+B=90^ [ ] (A+B)= 90^ + 1- =1 0 1- =0 =1 (1) Now + = + = + = (A+B) = 90^ . (90^ -A) =1 + =2 2 =2 2A (2) From (1) and (2) we get x^2- (2 2A )x+1=0 Hence, the value of the filler is x^2- (2 2A )x+1=0.

Fill in the blank. In a triangle ABC with =90^ the equation whose…

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Fill in the blank.
In a triangle ABC with $\angle\text{C}=90^\circ$ the equation whose roots are tan A and tan B is _______.
[Hint: $\text{A + B}=90^\circ\Rightarrow\tan\text{A}\tan\text{B}=1$ and $\tan\text{A}+\tan\text{B}=\frac{2}{\sin2\text{A}}$ ]

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Answer

In a triangle ABC with $\angle\text{C}=90^\circ$ the equation whose roots are tan A and tan B is $\text{x}^2-\Big(\frac{2}{\sin2\text{A}}\Big)\text{x}+1=0.$
Solution:
Given $\text{a}\triangle\text{ABC}$ with$\angle\text{C}=90^\circ$
$\text{x}^2-(\tan\text{A}+\tan\text{B})\text{x}+\tan\text{A}.\tan\text{B}=0$
$\text{A+B}=90^\circ[\because\angle\text{C}]$
$\Rightarrow\tan(\text{A+B})=\tan90^\circ$
$\Rightarrow\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}=\frac{1}{0}$
$\Rightarrow 1-\tan\text{A}\tan\text{B}=0$
$\Rightarrow\tan\text{A}\tan\text{B}=1\dots(1)$
$$Now $\tan\text{A}+\tan\text{B}=\frac{\sin\text{A}}{\cos\text{B}}+\frac{\sin\text{B}}{\cos\text{B}}$
$=\frac{\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}$
$=\frac{\sin\text{(A+B)}}{\cos\text{A}\cos\text{B}}=\frac{\sin90^\circ}{\cos\text{A}.\cos(90^\circ-\text{A})}$
$=\frac{1}{\cos\text{A}\sin\text{A}}$
$\therefore\tan\text{A}+\tan\text{B}=\frac{2}{2\sin\text{A}\cos\text{A}}=\frac{2}{\sin2\text{A}}\dots(2)$
$$From (1) and (2) we get
$\text{x}^2-\Big(\frac{2}{\sin2\text{A}}\Big)\text{x}+1=0$
Hence, the value of the filler is $\text{x}^2-\Big(\frac{2}{\sin2\text{A}}\Big)\text{x}+1=0.$

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