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Prove That: 2 Sin 68 ∘ Cos 10 ∘ − 2 Cot 15 ∘ 5 Tan 75 ∘ = ( 3 Tan 45 ∘ T a N 20 ∘ Tan 40 ∘ Tan 50 ∘ Tan 70 ∘ ) 5 = 1 - Mathematics

Sum

Prove that:

`(2  "sin"  68^circ)/(cos 10^circ )- (2  cot 15^circ)/(5 tan 75^circ) = ((3  tan 45^circ t  an 20^circ  tan 40^circ tan 50^circ tan 70^circ)) /5= 1` 

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Solution

LHS = `(2  "sin"  68^circ)/(cos 10^circ )- (2  cot 15^circ)/(5  tan 75^circ) = ((3  tan 45^circ tan 20^circ  tan 40^circ tan 50^circ tan 70^circ)) /5`

`=(2 sin 68^circ)/sin(90^circ - 22^circ) - (2  cot 15^circ)/(5 cot (90^circ - 75^circ)) - 3xx1xxcot(90^circ-20^circ)xxcot(90^circ-40^circ)xxtan 50^circxxtan 70^circ`

`= 2 - 2/5 = (3xx1/(tan 70^circ)xx1/(tan 50)^circxxtan 70^circ )/5`

`= 2 - 2/5 = 3/5`

`=( 10 - 2 - 3)/5`

`= 5/5`

 = 1

= RHS 

Concept: Trigonometry
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APPEARS IN

RS Aggarwal Secondary School Class 10 Maths
Chapter 7 Trigonometric Ratios of Complementary Angles
Exercise | Q 4.3 | Page 313
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