The value of $\frac{3cos^227°-5+3cos^263°}{tan^232°+4-cosec^258°} + sin 35°cos55° + cos35°sin55°$ is :
Answer & explanation
Correct answer: option 3
Using ,
sinA = cosB Iff A + B = 90 °
secA = cosecB Iff A + B = 90 °
$\frac{3cos^227°-5+3cos^263°}{tan^232°+4-cosec^258°} + sin 35°cos55° + cos35°sin55°$
= $\frac{3sin^263°-5+3cos^263°}{tan^232°+4-sec^232°} + sin 35°sin35° + cos35°cos35°$
= $\frac{3-5}{4-1} + 1 $
= $\frac{-2}{3} + 1 $
= $\frac{1}{3} $