The value of $\frac{cos29°cosec61°tan45°+2sin35°sec55°}{3sin^242°+3sin^248°}$ is:

1

$\frac{cos29°cosec61°tan45°+2sin35°sec55°}{3sin^242°+3sin^248°}$

Concept used :-

cosA = sinB   ( iff A + B = 90° )

 secA = cosecB  ( iff A + B = 90° ) 

And tan45° = 1

Now,

cosec61° = sec29°  & sec55° = cosec35°

=  \(\frac{cos29°.sec29°.tan45° + 2sin35°.cosec35°  }{3sin²42° + 3cos²42°}\)

=  \(\frac{1 + 2  }{3}\)

= 1