C-09-Q167mediumsingle_mcq1−sin245∘1+sin245∘+tan245∘=?\dfrac{1 - \sin^2 45^\circ}{1 + \sin^2 45^\circ} + \tan^2 45^\circ = ?1+sin245∘1−sin245∘+tan245∘=?a13\dfrac{1}{3}31b43\dfrac{4}{3}34c53\dfrac{5}{3}35d222ব্যাখ্যাSince sin245∘=12\sin^2 45^\circ=\tfrac12sin245∘=21 and tan245∘=1\tan^2 45^\circ=1tan245∘=1, the fraction becomes 1−121+12=1/23/2=13\frac{1-\frac12}{1+\frac12}=\frac{1/2}{3/2}=\tfrac131+211−21=3/21/2=31. Adding 111 gives 13+1=43\tfrac13+1=\tfrac4331+1=34.