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solve and post a solution soon..

Note by Dheeraj Agarwal 2 years, 5 months ago

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Yesss...., I also got it!! – Piyush Maheshwari · 2 years, 4 months ago

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\(Yess.. option is \boxed{a}\) – Mehul Chaturvedi · 2 years, 5 months ago

We know, tan(90) = [ tan(70) + tan(20) ] / [ 1 - tan(70).tan(20) ]. Therefore, 1 - tan(70).tan(20) = 0 So, tan(70).tan(20) = 1 ...(A)

Expanding tan(70) = [ tan(50) + tan(20) ] / [ 1- tan(50).tan(20) ]. Therefore, tan(50) + tan(20) = tan(70) * [ 1- tan(50).tan(20) ]. that is, tan(50) + tan(20) = tan(70) - tan(70).tan(50).tan(20). that is, 2.tan(50) + tan(20) = tan(50) + tan(70) - tan(70)tan(50).tan(20). = tan(50) + tan(70) - tan(50) ....[ from equation (A) ] = tan(70). – Mehul Chaturvedi · 2 years, 5 months ago

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## Comments

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TopNewestYesss...., I also got it!! – Piyush Maheshwari · 2 years, 4 months ago

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\(Yess.. option is \boxed{a}\) – Mehul Chaturvedi · 2 years, 5 months ago

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We know, tan(90) = [ tan(70) + tan(20) ] / [ 1 - tan(70).tan(20) ]. Therefore, 1 - tan(70).tan(20) = 0 So, tan(70).tan(20) = 1 ...(A)

Expanding tan(70) = [ tan(50) + tan(20) ] / [ 1- tan(50).tan(20) ]. Therefore, tan(50) + tan(20) = tan(70) * [ 1- tan(50).tan(20) ]. that is, tan(50) + tan(20) = tan(70) - tan(70).tan(50).tan(20). that is, 2.tan(50) + tan(20) = tan(50) + tan(70) - tan(70)tan(50).tan(20). = tan(50) + tan(70) - tan(50) ....[ from equation (A) ] = tan(70). – Mehul Chaturvedi · 2 years, 5 months ago

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