Question

using tan (a+b)=tana+tanb/1-tana tanb p.t tan75=(2+root3)​

Answer 1

Answer:

Step-by-step explanation:

From the properties of trigonometric ratios :

        tan45° = 1   ;    tan30° = 1 / √3

 Given,

        tan( a + b ) = ( tana + tanb ) / ( 1 - tana.tanb )

Here,

⇒ tan75°

  ⇒ tan( 45° + 30° )

  ⇒ ( tan45° + tan30° ) / ( 1 - tan45°tan30° )

  ⇒ { 1 + ( 1 / √3 ) } / { 1 - ( 1 * 1 / √3 ) }

  ⇒ { ( √3 + 1 ) / √3 } / { 1 - ( 1 / √3 ) }

  ⇒ { ( √3 + 1 ) / √3 } / { ( √3 - 1 ) / √3 }  

  ⇒ ( √3 + 1 ) / ( √3 - 1 )

 

Multiply as well as divide by ( √3 + 1 ):

⇒ ( √3 + 1 )( √3 + 1 ) / ( √3 - 1 )( √3 + 1 )

⇒ ( √3 + 1 )^2 / { ( √3 )^2 - ( 1 )^2 }

⇒ ( √3 + 1 )^2 / ( 3 - 1 )

⇒ ( 3 + 1 + 2√3 ) / 2

⇒ ( 4 + 2√3 ) / 2

⇒ 2 + √3

  Hence proved.

Answer 2

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

Using tan (a+b) = (tan a+tan b)/( 1-tan a tan b) , Prove that tan 75° = (2 + √3)

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{\orange{Given}}}}}$

• tan (a+b) = (tan a+tan b)/( 1-tan a. tan b ) ............(1)

$\Large{\underline{\mathfrak{\bf{\orange{To\:Prove}}}}}$

• tan 75° = (2 + √3)

$\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$

we calculate here, tan 75°

So,

➥ tan 75° = tan (45° + 30° )

Using first formula ,

➥ tan 75° = ( tan 45° + tan 30° )/ ( 1- tan 45°. tan 30°)

___________________

We know,

★ tan 45° = 1

★ tan 30° = 1/√3

____________________

Keep in above equation

➥tan 75° = (1 + 1/√3)/( 1- 1 × 1/√3}

➥tan 75° = ( √3 + 1) / ( √3 - 1)

Rationalize Denominator,

➥tan 75° = ( √3 + 1)(√3 + 1) /( √3 - 1)(√3 + 1)

➥tan 75° = [ ( √3)² + 1² + 2.1.√3 ] / [ (√3)² - (1)² ]

➥tan 75° = (3 + 1 + 2√3)/(3 - 1)

➥tan 75° = ( 4 + 2√3 ) / 2

Take common 2 in Numerator ,

➥tan 75° = 2 ( 2 + √3)/2

➥tan 75° = ( 2 + √3 )

That's prived.