Question

bhai koi ans do jaldi ​

Answer 1

cosec ( 90 - @ ) = sec @

Cot ( 90 - @ ) = tan @

sin ^2 ( 55 ) = cos ^2 ( 90 - 55 ) = cos ^2 ( 35 )

tan ( 70 ) = cot ( 90 - 70 ) = cot ( 20 )

tan ( 80 ) = cot ( 90 - 80 ) = cot ( 10 )

tan ( 60 ) = root ( 3 )

Also we know that

1 + tan ^ 2 ( @ ) = sec ^ 2 ( @ )

=> sce ^2 (@) - tan ^2 (@) = 1

Sin^2(@) + cos^2(@) = 1

=> cos^2(35 ) + sin^2 (35) = 1

Tan (@ ) × Cot (@) = 1

=> Tan (10) × Cot ( 10 ) = 1 and

tan (20 ) × cot ( 20 ) = 1

Now solution :

{ sec ( @ ) × sec ( @ ) - tan (@) × Cot (@) + Cos^ (35 ) + sin ^2 (@) } / [ { tan (10) × cot(10)} × { tan(20) × cot(20) } × tan (60) ]

=

[sec^2(@) - tan^2(@) + Cos^ (35 ) + sin ^2 (@) ]

/ ( 1 ) × ( 1 ) × ( root (3) )

= ( 1 + 1 ) / Root ( 3 ) --- Using above notes

= 2 / root (3)

Hope this help...sorry for poor representation....