Question

If tan A+ sin A = m and tan A - sin A = n, prove that:

(1) (m² - n²)² = 16mn
(2) m² - n² = 4√(mn)​

Answer 1

⧪ Given :

⊚ tan A + sin A = m

⊚ tan A - sin A = n

⧪ To Prove :

➀  m² - n² = 4√(mn)​

➁  (m² - n²)² = 16mn

⧪ Proof :

m = tan A + sin A

Squaring on both sides ,

➠ m² = (tan A + sin A)²

➠ m² = tan²A + sin²A + 2 tan A sin A

n = tan A - sin A

Squaring on both sides ,

➠ n² = (tan A - sin A)²

➠ n² = tan²A + sin²A - 2 tan A sin A

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Calculate m² - n²   ( LHS )

➙ tan²A + sin²A + 2 tan A sin A - (tan²A + sin²A - 2 tan A sin A)

➙ 4 tan A sin A

➙ $\sf 4\ \sqrt{(tan\ A.sin\ A)^2}$

➙ $\sf 4\ \sqrt{(tan^2A.sin^2A)}$

• sin²A = 1 - cos²A

➙ $\sf 4\ \sqrt{tan^2A(1-cos^2A)}$

➙ $\sf 4\ \sqrt{tan^2A-sin^2A}$

• a² - b² = (a + b)(a - b)

➙ $\sf 4\ \sqrt{(tan\ A+sin\ A)(tan\ A-sin\ A)}$

• m = tan A + sin A  [Given]
• n = tan A - sin A  [Given]

➙ $\sf 4\ \sqrt{(mn)}$  ( RHS )

So , m² - n² = 4√(mn) __ ➀

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Squaring on both sides ,

➤ (m² - n²)² = (4√(mn))²

➤ (m² - n²)² = 16mn __ ➁

$\orange{\bigstar}$  Hence proved  $\green{\bigstar}$

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Answer 2

Given : -

tan A + sin A = m

tan A - sin A = n

Required to prove : -

• ( m² - n² )² = 16mn

• m² - n² = 4√( mn )

Identity used : -

Sin² A + cos² A = 1

Solution : -

tan A + sin A = m

tan A + sin A = m tan A - sin A = n

we need to prove that ;

• ( m² - n² )² = 16 mn

• m² - n² = 4√( mn )

So,

Let's prove the 1st one ,

( m² - n² )² = 16 mn

Consider the LHS part

➜ ( m² - n² )²

➜ m² - n² = ( m + n ) ( m - n )

➜ ( [ m + n ] [ m - n ] )²

➜ ( [ tan A + sin A + tanA - sin A ] [ tan A + sin A - ( tan A - sin A ) ] )²

➜ [ ( tan A + sin A + tan A - sin A ) ( tan A + sin A - tan A + sin A ) ]²

➜ [ ( tan A + tan A ) ( sin A + sin A ) ]²

➜ [ 2 tan A 2 sin A ]²

➜ [ 4 tan A sin A ]²

➜ 16 tan² A sin² A

Consider the RHS part

16 mn

➜ 16 ( tan A + sin A ) ( tan A - sin A )

➜ 16 ( tan² A - sin² A )

[ From : ( a + b ) ( a - b ) = a² - b2 ]

➜ 16 ( sin²A/ cos²A - sin²A )

➜ 16 ( sin²A - sin²Acos²A/ cos²A )

Taking sin² A common

➜ 16 ( sin² A ( 1 - cos²A )/cos²A )

➜ 16 ( sin² A / cos² A ( sin²A ) )

From the identity sin² A + cos² A = 1 ,

sin² A = 1 - cos² A

➜ 16 ( sin²A / cos²A ( sin² A ) )

➜ 16 tan² A sin² A

LHS = RHS

Hence proved !

Let's prove the 2nd one ,

m² - n² = 4 √( mn )

Consider the LHS part

➜ m² - n²

➜ ( m + n ) ( m - n )

From the identity ;

( a² - b² ) = ( a + b ) ( a - b )

➜ ( tan A + sin A + tan A - sin A ) ( tan A + sin A - [ tan A - sin A ] )

➜ ( tan A + sin A + tan A - sin A ) ( tan A + sin A - tan A + sin A )

➜ ( 2 tan A ) ( 2 sin A )

➜ 4 tan A sin A

Consider the RHS Part

➜ 4 √ ( mn )

➜ 4 √ [ ( tan A + sin A ) ( tan A - sin A ) ]

➜ 4 √ [ tan² A - sin² A ]

From the identity ; .

( a + b ) ( a - b ) = a² - b²

➜ 4 √ [ sin² A / cos ² A - sin² A ]

➜ 4 √ [ sin² A - sin² A cos² A / cos² A ]

Taking sin² A as common

➜ 4 √ [ sin² A / cos² A ( 1 - cos² A ) ]

➜ 4 √ [ sin² A / cos² A ( sin² A ) ]

Since,

sin A / cos A = tan A

➜ 4√ [ tan² A sin² A ]

➜ 4 √ [ tan A sin A ]²

➜ 4 tan A sin A

LHS = RHS

Hence proved !