For A=133°, 2cosA2 is equal to
-(1+sinA)–(1–sinA)
-(1+sinA)+(1–sinA)
(1+sinA)–(1–sinA)
(1+sinA)+(1–sinA)
Explanation for correct option:
Step 1.Given that, A=133°
A=133°A2=66.5°
Step 2. using sin2A+cos2A=1;sin2A=2sinAcosA
1+sinA=(cosA2+sinA2)21+sinA=cosA2+sinA21+sinA=cosA2+sinA2.....(i)
Also,
1-sinA=(cosA2-sinA2)21-sinA=cosA2-sinA2[sincecosA2<sinA2]1-sinA=-cosA2+sinA2...(ii)
Step 3. Equation (i)-(ii), we get
1+sinA-1-sinA=cosA2+sinA2-(-cosA2+sinA2)1+sinA-1-sinA=2cosA2
Hence, correct option is (C)
11n+2+122n+1 is divisible by 133 for all nϵN.