Math, asked by saralanaidu6, 10 months ago

sin30+tan45-cosec60by cot45+cos60-sec60​

Answers

Answered by Anonymous
16

 \bf \huge \underline \purple{Answer : }

 \sf \implies \:  \frac{sin \: 30 + tan \: 45 -  \: cosec \: 60}{cot \: 45 + cos \: 60 - sec \: 60}  \\

 \sf \implies \:  \frac{ \frac{1}{2} + 1 -  \frac{2}{ \sqrt{3} }  }{ \frac{2}{ \sqrt{3}  } +  \frac{1}{2}  + 1 }  \\

 \sf \implies \:  \frac{ \sqrt{3} + 2 \sqrt{3}  - 4 }{4 +  \sqrt{3} + 2 \sqrt{3}  }  \\

 \sf \implies \:  \frac{3 \sqrt{3} - 4 }{4 + 3 \sqrt{3} }  \\

Rationalising the denominator,

 \sf \implies \:  \frac{3 \sqrt{3}  - 4}{4 + 3 \sqrt{3} }  \times  \frac{3 \sqrt{3}  - 4}{3 \sqrt{3} - 4 }  \\

 \sf \implies \:  \frac{ {(3 \sqrt{3}  - 4})^{2} }{ {(3 \sqrt{3} })^{2}  -  {(4)}^{2} }  \\

 \sf \implies \:  \frac{27 + 16 - 24 \sqrt{3} }{27 - 16}  \\

 \sf \implies \:  \frac{43 - 24 \sqrt{3} }{11}  \\

Answered by Anonymous
0

Step-by-step explanation:

Rationalising the denominator,

\begin{gathered}\sf \implies \: \frac{3 \sqrt{3} - 4}{4 + 3 \sqrt{3} } \times \frac{3 \sqrt{3} - 4}{3 \sqrt{3} - 4 } \\\end{gathered}

4+3

3

3

3

−4

×

3

3

−4

3

3

−4

\begin{gathered}\sf \implies \: \frac{ {(3 \sqrt{3} - 4})^{2} }{ {(3 \sqrt{3} })^{2} - {(4)}^{2} } \\\end{gathered}

(3

3

)

2

−(4)

2

(3

3

−4)

2

\begin{gathered}\sf \implies \: \frac{27 + 16 - 24 \sqrt{3} }{27 - 16} \\\end{gathered}

27−16

27+16−24

3

\begin{gathered}\sf \implies \: \frac{43 - 24 \sqrt{3} }{11} \\\end{gathered}

11

43−24

3

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