Question

the value of (1+tan32°)(1+tan13°)/(1+tan23°)(1+tan22°) is​

Answer 1

Answer:

Answer is 1

Step-by-step explanation:

(1+tan32)(1+tan13)= 1+tan32+tan13+tan32tan13 ∵(1+a)(1+b)=1+a+ab

(1+tan23)(1+tan22)=1+tan22+tan23+tan22tan23

32+13=45

apply tan on both sides

tan(32+13)=tan(45)

⇒ \frac{tan32+tan13}{1-tan32tan13}=tan45

1−tan32tan13

tan32+tan13

=tan45 ∵tan(A+B)=\frac{tanA+tanB}{1-tanAtanB}

1−tanAtanB

tanA+tanB

⇒tan32+tan13=1-tan32tan13

⇒tan13+tan32+tan32tan13=1

similarly 22+23=45

⇒tan22+tan23+tan22tan23=1

substituting values in above equations

\frac{(1+tan32)(1+tan13)}{(1+tan23)(1+tan22)}=\frac{1+1}{1+1}

(1+tan23)(1+tan22)

(1+tan32)(1+tan13)

=

1+1

1+1

=1

Answer 2

SOLUTION

TO DETERMINE

The value of

$\displaystyle \sf{ \frac{(1 + \tan {32}^{ \circ} )(1 + \tan {13}^{ \circ} )}{(1 + \tan {23}^{ \circ} )(1 + \tan {22}^{ \circ} )} }$

EVALUATION

First we find a general rule to find the required value of the given expression

Let A & B are two angles such that A + B = 45°

$\sf{A + B = {45}^{ \circ} }$

$\implies \sf{ \tan(A + B) = \tan {45}^{ \circ} }$

$\displaystyle\implies \sf{ \frac{\tan A + \tan B}{1 - \tan A \tan B\: } = 1}$

$\displaystyle\implies \sf{ \tan A + \tan B = 1 - \tan A \tan B }$

$\displaystyle\implies \sf{ \tan A + \tan B + \tan A \tan B = 1 }$

$\displaystyle\implies \sf{ 1 + \tan A + \tan B + \tan A \tan B = 2 }$

$\displaystyle\implies \sf{ (1 + \tan A ) + \tan B(1 + \tan A ) = 2 }$

$\displaystyle\implies \sf{ (1 + \tan A ) (1 + \tan B ) = 2 }$

Now given expression

$= \displaystyle \sf{ \frac{(1 + \tan {32}^{ \circ} )(1 + \tan {13}^{ \circ} )}{(1 + \tan {23}^{ \circ} )(1 + \tan {22}^{ \circ} )} }$

Numerator

$= \sf{ (1 + \tan {32}^{ \circ} )(1 + \tan {13}^{ \circ} )}$

Since 32° + 13° = 45°

So by the above formula

$\sf{ (1 + \tan {32}^{ \circ} )(1 + \tan {13}^{ \circ} )} = 2$

Denominator

$= \sf{ (1 + \tan {23}^{ \circ} )(1 + \tan {22}^{ \circ} )}$

Since 23° + 22° = 45°

So by the above formula

$\sf{ (1 + \tan {23}^{ \circ} )(1 + \tan {22}^{ \circ} )} = 2$

Hence the given expression

$= \displaystyle \sf{ \frac{(1 + \tan {32}^{ \circ} )(1 + \tan {13}^{ \circ} )}{(1 + \tan {23}^{ \circ} )(1 + \tan {22}^{ \circ} )} }$

$= \displaystyle \sf{ \frac{2}{2} }$

$= 1$

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