Question

how can we draw a maths figure in brainly for asking question and i know one option is there and can anyone plz show me that option​

Answer 1

Answer:

Step-by-step explanation:

what's the question ?

* this multiplication

+ this addition

- this subtraction

/ this division

α this alpha

β this beta

^ this raise to power

% this percentage

√ this root

Answer 2

Answer:

Step-by-step explanation:Tan42  Tan 78

= Tan (60 - 18) * Tan (60 + 18)

= (Tan 60 - tan 18) / (1+tan60 tan18)  * (tan60 + tan18) / (1 - Tan60 tan18)

= (√3 - tan18) (√3 + tan18) / [ (1+√3 tan18)(1-√3 tan18) ]

= (3 - tan²18) / (1 - 3 tan²18)      ---- (1)

=  (3 Cos²18 - Sin²18) / (Cos²18 - 3 Sin²18)

=  (3 - 4 Sin²18) / (4 Cos²18 - 3)

=  [ Sin 3*18 / Sin 18 ] / [ Cos 3*18 / Cos 18 ]

= Tan 54 / Tan 18                                       ---- (2)

      from the formula Sin 3A = sinA (3 - 4 sin²A)  and Cos3A = CosA (4 Cos²A - 3)

Tan 6 Tan 66

 = Tan (60 + 6) * tan 6 * tan (60-6) / tan 54  

 = tan (60 - 6) * tan (60 +6) * tan 6 / tan 54

 = [(√3 + tan 6) / (1 -√3 tan 6) ] * [(√3 + tan 6) / (1 - √3 tan6)] * tan 6 / tan 54

 = [ (3 - tan² 6) / (1 - 3 tan² 6) ] * tan 6 / tan 54

 = [ (3 Cos² 6 - Sin² 6) / ( Cos² 6 - 3 Sin² 6) ] * [ tan 6 / tan 54 ]

 = [ (3 - 4 Sin² 6) / (4 Cos² 6 - 3) ] * tan 6 / tan 54 ]

 = [ (3 Sin 6 - 4 Sin³ 6) / (4 Cos³ 6 - 3 Cos 6) ] / tan 54 ]

 = Sin (3*6) / Cos (3*6)  / tan 54

 = Tan 18 / Tan 54

Hence,  Tan 6 tan 42 tan 66  tan 78

   = Tan 54 / Tan 18  *  tan 18 / Tan 54  

   = 1

================================

Another way:

Tan (60 - A)

       = (tan 60 - tan A) / (1 + tan 60 tan A)

       = (√3 - Tan A) / (1 + √3 tan A)                    ---- (1)

Tan (60 +A)

       = (tan 60 + tan A) / (1 - Tan60 Tan A)

       = ( √3 + Tan A) / (1 - √3 Tan A)                  ---- (2)

Tan (60 - A) * Tan (60 + A)

       = (3 - Tan²A) / (1 - 3 Tan²A)

       = (3 - sec²A + 1) / (1 - 3 sec²A + 3)

       = (4 Cos²A -1) / (4 Cos²A - 3)              --- (3)