Question

solve the problem please .............​

Answer 1

Question :-

Prove That :- 4cos36° + cot7(1/2)° = √1 + √2 + √3 + √4 + √5 + √6

Solution :-

Taking LHS,

→ 4cos36° + cot(15/2)°

Lets Try to Solve Both Parts Seperately,

First Part :- 4cos36°

→ 4cos36°

→ 4cos(2*18°)

using cos2A = 1 - 2sin²A , we get,

→ 4[ 1 - 2sin²18° ]

Putting value of sin18° = (√5 - 1)/4 , we get,

→ 4 [ 1 - 2(√5-1/4)²]

→ 4[1 - 2{(5 + 1 - 2√5)/16}]

→ 4[1 - 2(6 - 2√5/16) ]

→ 4[ 1 - 2*2(3 - √5/16) ]

→ 4[ 1 - (3 - √5)/4 ]

→ 4 - 3 + √5

→ 1 + √5 ----------------- Equation (1).

___________________

Now, Solving second Part :- cot(15/2)°

→ cot(7.5°)

Using cotA = (cosA/sinA) ,

→ (cos7.5°/sin7.5°)

Mutliply & Divide by 2cos(7.5°) ,

→ (cos7.5°/sin7.5°) * {2cos(7.5°)/2cos(7.5°)}

→ (2cos²7.5°) / (2*sin7.5*cos7.5°)

using 2cos²A = 1 + cos2A in Numerator, & 2sinA*cosA = sin2A in denominator , we get,

→ [ (1 + cos15°) / sin15° ]

→ [ {1 + cos(45° - 30°)}/sin(45° - 30°) ]

Now, using

In Numerator :- cos(A - B) = cosA*cosB + sinA*sinB

in Denominator :- sin(A - B) = sinA*cosB - cosA*sinB

So,

→ Numerator = { 1 + (cos45°*cos30° + sin45°*sin30°) } = [ 1 + {(1/√2) * (√3/2) + (1/√2) * (1/2) ] = [ 1 + { (√3/2√2) + (1/2√2) }] = [ 1 + {(√3+1)/2√2} ] = [ (1 + √3 + 2√2) / (2√2) ]

And,

→ Denominator = sin45°*cos30° - cos45°*sin30° = (1/√2)*(√3/2) - (1/√2)(1/2) = (√3/2√2) - (1/2√2) = {(√3 - 1)/(2√2)}

Putting Both Values ,

→ [ (1 + √3 + 2√2) / (2√2) ] / [ {(√3 - 1)/(2√2)} ]

→ ( 1 + √3 + 2√2) / (√3 - 1)

Rationalizing Now, we get,

→ [ ( 1 + √3 + 2√2) / (√3 - 1) ] * [ (√3 + 1) / (√3 + 1) ]

→ [ ( 1 + √3 + 2√2) * (√3 + 1) / 2 ]

→ (1/2) [ √3 + 3 + 2√6 + 1 + √3 + 2√2 ]

→ (1/2) [ 1 + 3 + 2√2 + 2√3 + 2√6 ]

→ (1/2) [ 4 + 2√2 + 2√3 + 2√6 ]

→ (1/2) * 2 [ 2 + √2 + √3 + √6 ]

→ (2 + √2 + √3 + √6)

since 2 = √4

→ ( √2 + √3 + √4 + √6) ------------- Equation (2).

____________________

Now, Putting Both Values From Equation (1) & (2) in Question, we get ,

→ 4cos36° + cot(15/2)°

→ (1 + √5) + ( √2 + √3 + √4 + √6)

→ √1 + √2 + √3 + √4 + √5 + √6 ( since 1 = √1 .) = RHS .

✪✪ Hence Proved ✪✪

Answer 2

To Prove :-

→ 4 cos 36° + cot (7.5)° = √1 + √2 + √3 +√4 + √5 + √6 .

Solution :-

Taking LHS :-

→ 4 cos 36° + cot ( 7 .5)°

We'll solve it's both parts separately -

→ Cot (7.5)°

We know that cot can be written as cos/sin

→ cos 7.5° / sin 7.5°

Multiplying and dividing by 2 cos 7.5°

→ cos (7.5) × 2 cos (7.5) / sin (7.5) × 2 cos (7.5)

→ 2 cos ( 7.5)² / 2 sin (7.5) . cos (7.5)

( Sin 2A = 2 sin A . Cos A )

→ 2 cos ( 7.5)² / sin (2× 7.5)

→ 2 cos ( 7.5)² / sin ( 15 )

( Cos 2A = 2 cos²A - 1 → Cos 2A +1 = Cos²A )

→ Cos ( 15° ) + 1 / sin (15°)

15° can be written as 45° - 30°

→ Cos ( 45° - 30° ) + 1 / sin ( 45° - 30° )

Cos ( A -B) = Cos A .Cos B + Sin A . Sin B, and

Sin ( A - B ) = Sin A . Cos B + Cos A .Sin B

→ 1 + Cos 45° . Cos 30° + Sin 45° . Sin 30° / Sin 45° . Cos 30° - Cos 45° . Sin 30°

Cos 45° = 1/ √2 = Sin 45°

Cos 30° = √3/2

Sin 30° = 1/2

→ 1 + (1/√2× √3/2 ) + ( 1/√2 × 1/2) / ( 1/√2 × √3/2) - ( 1/√2 . 1/2 )

→ 2√2 + √3 + 1 / √3 - 1

Multiplying and dividing by √3 + 1

→ ( 2√2 + √3 + 1 ) × ( √3 +1)/ (√3 - 1 )×(√3+1)

→ 2√6 + 3 + √3 + 2√2 + √3 + 1 / 3-1

→ 2 ( √6 + √3 + √2 + √4 ) / 2

→ √ 2 + √3 + √4 + √6 . .......... Result 1st

Second part → 4 cos 36°

→ Cos 2x = 1 - 2 sin²x

Putting x here π/10 ie ( 18°)

→ Cos 2(18°) = 1 - 2 sin²(18)

And we know sin 18° =√5 -1 /4

→ Cos 36° = 1 - 2 ( √5 - 1 / 4 )²

(√5 )² + 1 - 2(√5) = ( √5 - 1 ) ²

→ 1 - 2 ( 6 - 2√5 /16 )

→ 1 - (3 -√5/4)

→ √5 + 1 / 4

4 cos 36° = 4 × ( √5 +√1 / 4)

→ √1 + √5 ......... Result 2nd

By Adding Both we got

4 cos 36° + cot 7.5 = √1 + √2 + √ 3 +√4+ √5 + √6

Hence proved .