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प्रश्न
Evaluate: `(2tan53°)/(cot 37°)-(cot 80°)/(tan 10°)`
बेरीज
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उत्तर
`(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
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Complimentary Angles for Tangent ( Tan ) and Contangency ( Cot )
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