Question

4cos2(this is square)A+4sinA=5, show that sec2(this is square)A−tan2(this is square)A=1

Answer 1

Solution :-

Given that ,

• 4cos²A + 4sinA = 5

We need to find ,

• sec²A - tan²A = 1

Simplifying the given ,

⇒ 4 [ cos²A + sinA ] = 5

⇒ cos²A + sinA = 5/4

⇒ [ 1 - sin²A ] + sinA = 5/4

⇒ sinA - sin²A = 5/4 - 1

⇒ sinA - sin²A = 1/4

⇒ sin²A - sinA + 1/4 = 0

Let sinA be x

⇒ x² - x + 1/4 = 0

Multiply with 4 on both sides ,

⇒ 4 [ x² - x + 1/4 ] = 0 × 4

⇒ 4x² - 4x + 1 = 0

Substituting ,

• - 4x = - 2x - 2x

⇒ 4x² - 2x - 2x + 1 = 0

⇒ 2x ( 2x - 1 ) - 1 ( 2x - 1 ) = 0

⇒ ( 2x - 1 ) ( 2x - 1 ) = 0

⇒ ( 2x - 1 )² = 0

⇒ 2x - 1 = 0

⇒ 2x = 1

⇒ x = 1/2

So , here x = sinA

⇒ sinA = 1/2

Substituting , 1/2 = sin30°

⇒ sinA = sin30°

By comparing ,

⇒ A = 30°

Now , proving sec²A - tan²A = 1

By substituting , A = 30°

➵ sec²30° - tan²30° = 1

➵ [ 2/√3 ] ² - [ 1/√3 ]² = 1

➵ 4/3 - 1/3 = 1

➵ 4 - 1/3 = 1

➵ 3/3 = 1

➵ 1 = 1

LHS = RHS

Hence , proved !!

Answer 2

Given:

• 4cos²A + 4sinA = 5

To Prove:

• sec²A - tan²A = 1

Solution:

Given that,

✒ 4cos²A + 4sinA = 5

➡ 4(cos²A + sinA )= 5

➡ (1 - sin²A) + sinA = 5/4

➡ sin²A - sinA = 5/4 - 1

➡ sin²A - sinA = 1/4

➡ sin²A - sinA + 1/4 = 0

[ Now, let's sinA = R ]

➡ R² - R + 1/4 = 0

[ Multiply by 4 both sides, we get ]

➡ 4R² - 4R + 1 = 0

➡ (2R - 1)² = 0

➡ 2R - 1 = 0

➡ R = 1/2

✒ So, sinA = 1/2 [ °.° sinA = R ]

➡ sinA = sin30° [ °.° 1/2 = sin30° ]

[ By comparing ]

➡ A = 30°

Now, Proving sec²A - tan²A = 1,

➡ sec²30° - tan²30° = 1 [ °.° A = 30° ]

➡ (2/√3)² - (1/√3)² = 1

➡ 4/3 - 1/3 = 1

➡ 3/3 = 1

➡ 1 = 1

LHS = RHS.

Hence Proved!!