Question

please answer this question.!​

Answer 1

Given :-

• m = tanA + sin A
• n = tanA - sin A

To Prove :-

• $\sf{{m}^{2} - {n}^{2} = 4 \sqrt{mn}}$

Solution

Taking LHS Firstly ,

→ LHS = m² - n²

We know that a²-b² = (a+b)(a-b) , Using this identity

→ m²-n² = (m+n)(m-n)

Putting the values of m and n here

→ $\sf{ (\tan A+ \sin A + \tan A- \sin A)(\tan A+ \sin A - \tan A+ \sin A)}$

→ $\sf{(2 \tan A)(2 \sin A) = 4 \tan A. \sin A }$

→ LHS = 4 tanA sinA

Now taking the RHS .

→ $\sf{ 4 \sqrt{mn} }$

Here also putting the value of m and n .

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

Here also we'll use the identity a²-b² = (a+b)(a-b)

→ $\sf{4 \sqrt{ ({ \tan }^{2} A ) - ({\sin}^{2} A) }}$

Now writing tan theta in form of sin and cos.

→ $\sf{ 4 \sqrt{{\frac{\sin A}{\cos A}}^{2} - {\sin}^{2} A } }$

Taking sin common from here .

→ $\sf{4 \sqrt{ {\sin}^{2} A ( \frac{1 - {\cos}^{2}A }{{\cos}^{2} A } ) }}$

Now we know that 1 - cos²A = sin²A.

→ $\sf{4 \sqrt{ {\sin}^{2}A . \frac{{\sin}^{2} A}{{\tan}^{2} A } } }$

→ $\sf{ 4 \sqrt{{\sin}^{2}A .{\tan}^{2}A } }$

→ RHS = 4 sinA. tanA

So LHS = RHS = 4 sinA. tanA .

Hence proved .

Answer 2

GIVEN:

• tanA + sinA = m ----------( 1 )
• tanA - sinA = n -------- ( 2 )

TO PROVE:

• m² - n² = 4√mn

SOLUTION:

Taking equation 1 & squaring on both sides

(tanA + sinA)² = m²

Assuming as equation 3

Taking equation 2 & squaring on both sides

(tanA - sinA)² = n²

Assuming as equation 4

Now,

Equation 3 - Equation 4

→ (tanA + sinA)² - (tanA - sinA)² = m² - n²

Using algebraic identity

(a + b)² - (a - b)² = 4ab

→ 4tanA.sinA

m² - n² = 4√mn = 4tanA.sinA

Taking ‘ 4√mn ’

Substituting the values of m & n

→ 4√(tanA + sinA)(tanA - sinA)

Simplifying using algebraic identity

(a + b)(a - b) = a² - b²

→ 4√(tan²A - sin²A)

We know that

tanA = sinA/cosA

→ 4√(sin²A/cos²A - sin²A)

→ 4√(sin²A - sin²A.cos²A/cos²A)

→ 4√[sin²A(1 - cos²A)/cos²A]

→ 4√[sin⁴A/cos²A]

→ 4√[sin²A × sin²A/cos²A]

→ 4√[sinA × tanA]²

→ 4sinA.tanA

Now comparing with ‘4tanA.sinA’

4sinA.tanA = 4tanA.sinA

LHS = RHS

Hence proved