Correct Answer – Option 2 : 1

Given:

\(\frac{{\cos \,{{29}^ \circ }\cos ec\,{{61}^ \circ }\tan \,{{45}^ \circ } + 2\sin \,{{35}^ \circ }\sec \,{{55}^ \circ }}}{{3{{\sin }^2}\,{{42}^ \circ } + 3{{\sin }^2}\,{{48}^ \circ }}}\)

Trigonometry properties:

cosecθ = 1/sinθ 

sin(90 – θ ) = cosθ 

secθ = 1/cosθ 

cos(90 – θ ) = sinθ 

sin²θ + cos²θ = 1

tan 45^o = 1

Calculation:

\(\frac{{\cos \,{{29}^ \circ }\cos ec\,{{61}^ \circ }\tan \,{{45}^ \circ } + 2\sin \,{{35}^ \circ }\sec \,{{55}^ \circ }}}{{3{{\sin }^2}\,{{42}^ \circ } + 3{{\sin }^2}\,{{48}^ \circ }}}\)

According to the trigonometry properties 

⇒ {cos 29^o (1/sin 61^o) tan 45^o + 2 sin 35^o (1/cos 55^o)}/ 3sin² 42^o + 3sin²(90^o – 48^o)

⇒ {(cos 29^o/cos61^o)tan 45^o + 2 sin 35^o/cos 55^o}/3(sin 42^o + cos42^o)

⇒ {(cos29^o/sin(90 ^o – 61^o))× 1 + 2 sin 35^o/cos(90^o – 55^o)}/3 × 1

⇒ {cos 29²/cos 29² + 2 sin 35^o/sin 35^o}/3

⇒ (1 + 2)/3

⇒ 1

∴ The value of \(\frac{{\cos \,{{29}^ \circ }\cos ec\,{{61}^ \circ }\tan \,{{45}^ \circ } + 2\sin \,{{35}^ \circ }\sec \,{{55}^ \circ }}}{{3{{\sin }^2}\,{{42}^ \circ } + 3{{\sin }^2}\,{{48}^ \circ }}}\) is 1.