The value of $\frac{cos29°cosec61°tan45°+2sin35°sec55°}{3sin^242°+3sin^248°}$ is:
Answer & explanation
Correct answer: option 4
$\frac{cos29°cosec61°tan45°+2sin35°sec55°}{3sin^242°+3sin^248°}$
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