Question

11 .. question plzz do it

Answer 1

Answer :-

1 / 3

Explanation :-

To solve it easily let us solve first term and then second

First term

[ sec² ( 90 - θ ) - cot² θ ] / [ 2( sin² 25° + sin² 55°) ]

Consider numerator

sec² ( 90 - θ ) - cot² θ

= cosec² θ - cot² θ

[ ∵ sec ( 90 - θ ) = cosec θ ]

= 1

[ ∵ cosec² θ - cot² θ = 1 ]

Consider denominator

2( sin² 25° + sin² 55 )

= 2( sin² 25° + sin² (90 - 25)° )

= 2( sin² 25° + cos² 25° )

[ ∵ sin ( 90 - θ ) = cos θ ]

= 2( 1 )

[ ∵ sin² θ + cos² θ = 1 ]

= 2

Hence, 1st term = Numerator / Denominator = 1 / 2

Second term

[ 2cos² 60°. tan² 28°. tan² 62° ] / [ 3 ( sec² 43° - cot² 47° ) ]

Consider numerator

2cos²60° . tan² 28° . tan² 62°

= 2cos² 60° . tan² 28° . tan² ( 90 - 28 )°

= 2cos² 60 . tan² 28° . cot² 28°

[ ∵ tan ( 90 - θ ) = cot θ ]

= 2cos² 60 . tan² 28 . ( 1 / tan² 28° )

[ ∵ cot θ = 1 / tan θ ]

= 2cos² 60 . 1

= 2cos² 60

= 2 ( 1 / 2 )²

[ ∵ cos 60° = 1 / 2 ]

= 2 ( 1 / 4 )

= 1 / 2

Consider denominator

3( sec² 43° - cot² 47° )

= 3( sec² 43° - cot² ( 90 - 43 )° )

= 3( sec² 43° - tan² 43° )

[ ∵ cot ( 90 - θ )° = tan θ ]

= 3( 1 )

[ ∵ sec² θ - tan² θ = 1 ]

= 3

Hence, 2nd term = Numerator / Denominator = ( 1 / 2 ) / 3 = ( 1 / 2 ) * ( 1 / 3 ) = 1 / 6

Now, [ sec² ( 90 - θ ) - cot² θ ] / [ 2( sin² 25° + sin² 55°) ] - [ 2cos² 60°. tan² 28°. tan² 62° ] / [ 3 ( sec² 43° - cot² 47° ) ]

= ( 1 / 2 ) - ( 1 / 6 )

= (3 - 1) / 6

= 2 / 6

= 1 / 3