Math, asked by snigdhapoddar123, 9 months ago

pls anyone help me for this sum​

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Answered by karannnn43
10

STEP BY STEP EXPLANATION :-

2( { \sec }^{2}35 -  { \cot }^{2}55) -  \frac{ \cos(28). \csc(62) }{ \tan(18). \tan(36). \tan(30) \tan(54). \tan(72)     }

 = 2( { \sec }^{2}35 -  { \cot }^{2}55) -  \frac{ \cos(28). \csc(90 - 28) }{ \tan(90 - 72). \tan(90 - 54). \tan(30) \tan(54). \tan(72)     }

  • Changing some terms into their complementary angles , we get ,

2( { \sec }^{2}35 -  { \cot }^{2}(90 - 35) -  \frac{ \cos(28). \sec(28) }{ \cot(72). \cot(54). \tan(30) \tan(54). \tan(72)     }

  • Since, cotA x tanA = 1 and secA x cosA = 1 , So, we get,

2( { \sec }^{2}35 -  { \tan}^{2}35)-  \frac{ 1 }{ \tan(30)    }

  • putting tan²= sec²A-1, we get.

2( { \sec }^{2}35 -{ \sec}^{2}35+1)-  \frac{ 1 }{ \tan(30)    }

(2 \times   1) -  \frac{1}{ \tan(30) }

  • Now, write the value of tan30

  2 -(1 \div  \frac{1}{ \sqrt{3} } )

 =   (2 -  \sqrt{3} )

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1

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