Question

if tan teeta +sin teeta and tan teeta-sin teeta= n then show that m2-n2=4√mn

Answer 1

Hi,

tanθ + sinθ = m -----------( 1 )

tanθ - sinθ = n ------------( 2 )

i ) m^2 - n^2

= ( tanθ + sinθ )^2 - ( tanθ - sinθ )^2

= 4tanθ sinθ ----( 3 )

ii ) 4 sqrt (mn)

= 4 sqrt[( tanθ + sinθ )( tanθ - sinθ )]

= 4 sqrt[ tan^2θ - sin^2θ ]

= 4 sqrt [ sin^2θ / cos^θ - sin^2 θ /1 ]

= 4 sqrt [ sin^2θ ( 1/ cos^2θ - 1/1 ) ]

= 4 sqrt [ sin^2 θ ( sec2 θ - 1) ]

=4 sqrt [ sin^2 θ tan^2θ ]

= 4 sinθ tanθ ------( 4 )

from ( 3 ) and ( 4 )

( 3 ) = ( 4 )

Therefore,

m^2 - n^2 = 4 sqrt( mn)


i hope this will useful to you.

*****

Answer 2

Question:

If tanθ + sinθ = m and tanθ - sinθ =n, Show that m² - n² = 4√mn

Answer:

First simplify the LHS, i.e, m² - n².

m² - n² = (tanθ + sinθ)² - [(tanθ - sinθ)²]

= tan²θ + sin²θ + 2tanθsinθ - [tan²θ + sin²θ - 2tanθ × sinθ]

= tan²θ + sin²θ + 2tanθsinθ - tan²θ - sin²θ + 2tanθ × sinθ

⇒ tan²θ + sin²θ and - tan²θ - sin²θ get cancelled

= 2tanθsinθ + 2tanθsinθ

= 4 × tanθ × sinθ → 1

Now, we simplify the RHS i.e, 4√mn

$\sf 4\sqrt{mn} = 4\sqrt{(tan\theta + sin\theta)(tan\theta - sin\theta)}$

$\boxed{\sf Using \ (a -b)(a + b) \ = \ a^2 \ - \ b^2}$

$\sf 4\sqrt{mn} = 4\sqrt{tan^2\theta - sin^2\theta}$

$\boxed{\sf Using \ tan^{2}\theta = \frac{sin^{2}\theta}{cos^2\theta}}$

$\sf 4\sqrt{mn} = 4\sqrt{\dfrac{sin^2\theta}{cos^2\theta} - sin^2\theta}$

$\sf 4\sqrt{mn} = 4\sqrt{\dfrac{sin^2\theta - cos^2\theta \times sin^2\theta}{cos^2\theta}}$

$\sf 4\sqrt{mn} = 4\sqrt{sin^2\theta \times {\dfrac{(1 - cos^2\theta)}{cos^2\theta}}$

$\boxed{\sf Using \ sin^2\theta \ = 1 \ - \ cos^2\theta}$

$\sf 4\sqrt{mn} = 4\sqrt{sin^2\theta \times {\dfrac{sin^2\theta}{cos^2\theta}}$

$\boxed{\sf Using \ tan^{2}\theta = \frac{sin^{2}\theta}{cos^2\theta}}$

$\sf 4\sqrt{mn} = 4\sqrt{sin^2\theta \times tan^2\theta$

$\large\text{\sf Squares and roots are cancelled.}$

= 4 × sinθ × tanθ → 2

From 1 and 2,

LHS = RHS,

Hence Proved!