1.

+sec²40°-cot²50°)/(sin²40°+sin²50°)-2cos60°sin20°sec70°​

Answer»

ANSWER:

\frac{ {sec}^{2}40 - {cot}^{2}50 }{ {sin}^{2}40 + {sin}^{2}50 } - 2 \cos(60) \sin(20) \sec(70) =

\frac{ {cosec}^{2}50 - {cot}^{2} 5 0| }{ {sin}^{2}40 + {cos}^{2}40 } - 2 \times \frac{1}{2} \times \sin(20) \csc(20) =

\frac{1}{1} - 2 \times \frac{1}{2} \times 1 = 1 - 1 = 0 \: \: ans.


Discussion

No Comment Found

Related InterviewSolutions