HOW TO FIND THE VALUES OF TAN 75 AND COT 75

How to Find the Values of Tan 75 and Cot 75 :

Here we are going to see how to find the values of tan 75 degree and cot 75 degree and some example problems based on those values.

How to Find the Values of Tan 75 and Cot 75 ?

Question 1 :

Show that tan 75° + cot 75° = 4.

Solution :

tan 75°  =  tan (45° + 30°)

  =  (tan 45° + tan 30°)/ (1 - tan 45° tan 30°)

  =  (1 + (1/√3)) /(1 - 1(1/√3))

  =  [(√3 + 1)/√3] / [(√3 - 1)/√3]

  =  (√3 + 1)/(√3 - 1)

Multiply by (√3 + 1) on both numerator and denominator.

  =  (√3 + 1)²/(√3² - 1²)

  =  (3 + 1 + 2√3) / (3 - 1)

  =  (4 + 2√3) / 2

  =  2 + √3   ------(1)

tan 15°  =  cot (90° - 15°)

tan 15°  =  cot 75°

Instead of finding the value of cot 75, let us find the value of tan 15.

tan 15°  =  tan (45° - 30°)

Using compound angle formula, we get

  =  (√3 - 1)/(√3 + 1)

Multiply by (√3 - 1) on both numerator and denominator.

  =  (√3 - 1)²/(√3² - 1²)

  =  (3 + 1 - 2√3) / (3 - 1)

  =  (4 - 2√3) / 2

  =  2 - √3   ------(2)

(1) + (2)

tan 75° + cot 75°  =    2 + √3  +  2 - √3  

tan 75° + cot 75°  =  4

Hence proved.

Question 2 :

Prove that cos(A + B) cosC−cos(B + C)cosA = sinB sin(C−A).

Solution :

L.H.S

  =  cos(A + B) cosC−cos(B + C)cosA

  =  (cosAcosB - sinAsinB)cosC - (cosBcosC - sinBsinC)cosA

  =  cosA cosB cosC - sinA sinB cosC - cosB cosC cosA + sinB sinC cosA

  =   sinB (sin C cos A - sin A cos C)

  =  sin B sin (C- A)

Hence proved.

Question 3 :

Prove that sin(n + 1)θ sin(n − 1)θ + cos(n + 1)θ cos(n − 1)θ = cos2θ, n ∈ Z.

Solution :

L.H.S

   =  sin(n + 1)θ sin(n − 1)θ + cos(n + 1)θ cos(n − 1)θ

  =   cos(n − 1)θ cos(n + 1)θ + sin(n − 1)θ sin(n + 1)θ

  =  cos [n - 1 - (n + 1)]θ

 =  cos  [n - 1 - n - 1)]θ

  =  cos (-2θ)

  =  cos 2θ  ------->  R.H.S

Hence proved. 

After having gone through the stuff given above, we hope that the students would have understood "How to Find the Values of Tan 75 and Cot 75". 

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