Question

Solve the ques in the attachment

Class 10

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Answer 1

Hey!!!!

Good Afternoon

Difficulty Level : Average

Chances of being asked in Board : 60%

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We will use the concept of complimentary angles in this question and also tan x cot = 1

Let thetha = x

So we have to find

= (- tanx.cot(90-x) + secx.cosec(90-x) + sin²35 + sin²55)/(tan10.tan20.tan30.tan80.tan70)

Here cot(90-x) = tanx
cosec (90-x) = secx
sin²35 = cos²55 (cosx = sin(90-x) )
tan80 = cot10
tan70 = cot20
tan30 = 1/√3

Replacing all

= √3(- tan²x + sec²x + cos²55 + sin²55)/tan10tan20cot10cot20

= √3(1 + 1)/1

= 2√3 <<<<<<< Answer

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Hope this helps ✌️

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Jai Hind

Answer 2

Answer :

Now,

$\frac{-tan \theta \: cot(90 - \theta) \: + sec \theta \: cosec(90 - \theta) + {sin}^{2} 35 + {sin}^{2}55 }{tan10 \: . \: tan20 \: . \: tan30 \: . \: tan70 \: . \: tan80} \\ \\ = \frac{-tan \theta \: tan \theta + sec \theta \: sec \theta + {sin}^{2} 35 + {sin}^{2}(90 - 35) }{tan10 \: . \: tan20 \: . \: tan30 \: .tan(90 - 20) \: . \: tan(90 - 10)} \\ \\ - - - - - - - - - - \\ tan(90 - \theta) = cot \theta \: \: \: \: \: \: | \\ cosec(90 - \theta) = sec \theta \: \: |\\sin(90 - \theta) = cos \theta \:\:\:\: \: |\\ - - - - - - - - - - \\ \\ = \frac{ - {tan}^{2} \theta + {sec}^{2} \theta + ({sin}^{2} 35 + {cos}^{2} 35) }{tan10 \: . \: tan20 \: . \: tan30 \: . \: cot20 \: . \: cot10} \\ \\ = \frac{ 1 + 1}{(tan10 \: . \: cot10).(tan20 \: . \: cot20).tan30} \\ \\ - - - - - - - - - \\ {sin}^{2} \theta + {cos}^{2} \theta = 1 \: \: \: \: | \\ {sec}^{2} \theta - {tan}^{2} \theta = 1 \: \: \: \: | \\ - - - - - - - - - \\ \\ = \frac{2 }{(1).(1).tan30} \\ \\ = \frac{2}{ \frac{1}{ \sqrt{3} } } \\ \\ - - - - - - - \\ tan30 = \frac{1}{ \sqrt{3} } \: \: \: \: \: | \\ - - - - - - - \\ = 2 \sqrt{3}$

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