Question

"Question17
Show that : 1 - sin60°/cos60° = tan60°-1/tan60° + 1
Chapter6,T-Ratios of particular angles Exercise -6 ,Page number 288"

Answer 1

Hey there!

We know that,
sin60 = √3/2
cos60 = 1/2
tan60 = √3

Now,
Finding the value of L. H. S,

1 - sin60/ ( cos60 )

=[ 1 - √3/2 ] / 1/2

= [ 2 - √3 / 2 ] * 2

= 2 - √3

Now, Finding Value of R. H. S,

tan60 - 1 / tan60 + 1

= √3 - 1 / √3 + 1 [ Rationalise denominator ]

= (√3 - 1 )² / 2

= 3 + 1 - 2√3 /2

= 4 - 2√3 /2

= 2 [ 2 - √3 ] /2

= 2 - √3 .

We see that, L. H. S = R. H. S = 2 - √3 .

Hence proved that, 1 - sin60°/cos60° = tan60°-1/tan60° + 1

Answer 2

HELLO DEAR,

we know that:-


sin60° = √3/2
cos60° = 1/2
tan60° = √3


NOW,


[1 - sin60°]/cos60°

= [1 - (√3/2) ]/ (1/2)

= [ ( 2 - √3)/2] / (1/2)

= (2 - √3) (R.H.S)



(tan60° - 1) / (tan60° + 1)


= (√3 - 1) / (√3 + 1)

= (√3 - 1) / (√3 + 1) × (√3 - 1)/(√3 - 1)

= (3 + 1 - 2√3)/ (3 - 1)

= (4 - 2√3)/2

= 2(2 - √3)/2

= (2 - √3) ( L.H.S)


HENCE,

R.H.S = L.H.S



I HOPE ITS HELP YOU DEAR,
THANKS