Math, asked by mehakbajwa002, 7 months ago

tan square60+ 4cos square45sec square30+5 cos square 90/cosec30 +sec60-cotsquare30​

Answers

Answered by swapnilnagargoje7499
0

Answer:tan square60+ 4cos square45sec square30+5 cos square 90/cosec30 +sec60-cotsquare30​

tan square60+ 4cos square45sec square30+5 cos square 90/cosec30 +sec60-cotsquare30​

Step-by-step explanation:tan square60+ 4cos square45sec square30+5 cos square 90/cosec30 +sec60-cotsquare30​

tan square60+ 4cos square45sec square30+5 cos square 90/cosec30 +sec60-cotsquare30​

tan square60+ 4cos square45sec square30+5 cos square 90/cosec30 +sec60-cotsquare30​

Answered by mathdude500
0

Answer:

Using trigonometric values of standard angles

Step-by-step explanation:

 { (\sqrt{3} )}^{2}  + 4 \times  \frac{1}{ {( \sqrt{2} )}^{2} }  \times ( { \frac{2}{ \sqrt{3} } )}^{2}  + 0 + 2 - ( { \sqrt{3}) }^{2}   \\ = 3  + \frac{8}{3}  + 2 - 3 \\  =  \frac{14 }{3}

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