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Tamil Nadu Board of Secondary EducationHSC Commerce Class 11

Prove that: tan(-225°) cot(-405°) – tan(-765°) cot(675°) = 0. - Business Mathematics and Statistics

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Question

Prove that:

tan(-225°) cot(-405°) – tan(-765°) cot(675°) = 0.

Sum

Solution

tan(-225°) = -(tan 225°)

= -(tan(180° + 45°))

= – tan 45°

= – 1

cot(-405°) = -(cot 405°)

= – cot(360° + 45°)     ....[∵ For 360° + 45° no change in T-ratio.]

= -cot 45°

= -1

tan(-765°) = -tan 765°

= -tan(2 × 360° + 45°)

= -tan 45°

= -1

cot 675° = cot (360°+ 315°)

= cot 315°

= cot(360° – 45°)

= -cot 45°

= -1

LHS = tan(-225°) cot(-405°) – tan(-765°) cot(675°)

= (-1) (-1) – (-1) (-1)

= 1 – 1

= 0

= RHS.

Hence proved.

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Trigonometric Ratios
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Chapter 4: Trigonometry - Exercise 4.1 [Page 81]

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Samacheer Kalvi Business Mathematics and Statistics [English] Class 11 TN Board
Chapter 4 Trigonometry
Exercise 4.1 | Q 5. (i) | Page 81
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