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. 

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