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Evaluate: 2 Tan 53 ° Cot 37 ° − Cot 80 ° Tan 10 ° - Mathematics

Sum

Evaluate: `(2tan53°)/(cot 37°)-(cot 80°)/(tan 10°)`

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Solution

`(2tan53°)/(cot 37°)-(cot 80°)/(tan 10°)`

= `(2tan(90°- 37°))/(cot 37°) - (cot (90° - 10°))/(tan 10°)`

= `(2cot 37°)/(cot 37°) - (tan 10°)/(tan 10°)`
= 2 - 1
= 1

Concept: Complimentary Angles for Tangent ( Tan ) and Contangency ( Cot )
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APPEARS IN

Selina Concise Mathematics Class 9 ICSE
Chapter 25 Complementary Angles
Exercise 25 | Q 2.4 | Page 310
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