Math, asked by palakbaliyan16, 9 months ago

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

Answers

Answered by rajeevr06
2

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.

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