Question

tan41-tan11/1+tan41-tan11​

Answer 1

SOLUTION

TO DETERMINE

$\displaystyle \sf{ \frac{ \tan {41}^{ \circ} - \tan {11}^{ \circ} }{1 + \tan {41}^{ \circ} \: \tan {11}^{ \circ}} }$

FORMULA TO BE IMPLEMENTED

We are aware of the Trigonometric identity that

$\displaystyle \sf{ \tan( \alpha - \beta ) = \frac{ \tan \alpha - \tan \beta }{1 + \tan \alpha \: \tan \beta } }$

EVALUATION

$\displaystyle \sf{ \frac{ \tan {41}^{ \circ} - \tan {11}^{ \circ} }{1 + \tan {41}^{ \circ} \: \tan {11}^{ \circ}} }$

Now we take

$\sf{ \alpha = {41}^{ \circ} \: \: \: and \: \: \beta = {11}^{ \circ} }$

So we get from above mentioned formula

$\displaystyle \sf{ \frac{ \tan {41}^{ \circ} - \tan {11}^{ \circ} }{1 + \tan {41}^{ \circ} \: \tan {11}^{ \circ}} }$

$\displaystyle \sf{ = \frac{ \tan \alpha - \tan \beta }{1 + \tan \alpha \: \tan \beta } }$

$\displaystyle \sf{ = \tan( \alpha - \beta ) }$

$\displaystyle \sf{ = \tan( {41}^{ \circ} - {11}^{ \circ} ) }$

$\displaystyle \sf{ = \tan {30}^{ \circ} }$

$\displaystyle \sf{ = \frac{1}{ \sqrt{3} } }$

FINAL ANSWER

$\boxed{ \: \: \displaystyle \sf{ \frac{ \tan {41}^{ \circ} - \tan {11}^{ \circ} }{1 + \tan {41}^{ \circ} \: \tan {11}^{ \circ}} } \: = \frac{1}{ \sqrt{3} } \: }$

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. prove that (1+cos18)(1+cos54)(1+cos126)(1+cos162)=1÷16

https://brainly.in/question/29235086

2. prove that :

tan9A - tan6A -tan3A = tan9A.tan 6A.tan tan 3A

https://brainly.in/question/9394506