Question

Prove that. Tan alpha +cotA=2cosec2A & deduce that tan75°+cot75°=4

Answer 1

Proof :

L.H.S. = tanA + cotA

= sinA/cosA + cosA/sinA

= (sin²A + cos²A)/(sinA cosA)

= 1/(sinA cosA) [∵ sin²A + cos²A = 1 ]

= 2/(2 sinA cosA)

= 2/sin2A [∵ 2 sinA cosA = sin2A ]

= 2 cosec2A

= R.H.S. [Proved]

Answer :

Now, tan75° + cot75°

= 2 cosec (2 × 75°)

= 2 cosec150°

= 2 cosec(90° + 60°)

= 2 sec60° [ ∵ cosec (90° + A) = secA ]

= 2 × 2

= 4 [Proved]

Hope it helps! (:

Answer 2

There is the little error in the question, Instead of the tan alpha it will be tan A. Thus we need to prove,  
     tan A + cot A=2 Cosec 2A 

From L.H.S., 

   tan A + cot A = Sin A/Cos A + Cos A/ Sin A 
    = (Sin² A + Cos² A)/Sin A CosA    [Taking L.C.M.] 
    = 1/SinACosA        [∵ Sin² A + Cos² A = 1]
    = 2/2 SinA CosA
    = 2/Sin 2A    [∵ Sn 2A = 2 Sin A Cos A] 
    = 2 × Cosec 2A  [∵ Sin 2A = 1/Cosec 2A]
    = 2 Cosec 2A 
    = R.H.S. 

∵ L.H.S. is proved to be equal to R.H.S. 
∴ Relation is Proved. 


Now, 
tan 75° + cot 75° = tan(30 + 45) + cot (90 - 15) 
   = tan (30 + 45) + tan 15
   = tan (45 + 30) + tan ( 45 - 30)

Now, using the formula, 
 tan (A + B)= tan A +tan B/(1 - tanA tanB)
tan (A - B) = tan A - tan B/(1 + tan AtanB) 
  
       ∴ tan 75° + cot 75° = (tan45° + tan 30°)/(1 - tan45°tan30°) + (tan 45° - tan30°)/(1 + tan45°tan30°)
    = (1 + 1/√3)/(1 - 1/√3) + (1 - 1/√3)/(1 + 1/√3)
    = [(1 + 1/√3)² + (1 - 1/√3)²]/[(1)² - (1/√3)²]
    = [1 + 1/3 + 2/√3 + 1 + 3 - 2√3]/[1 - 1/3]
    = [2 + 2/3]/[1 - 1/3]
    = [6 + 2]/[2]
    = 8/2 
    = 4 


Hence Proved. 


Hope it helps.