Write all trigonometric ratios of angle A in terms of sec A

1.	Write all trigonometric ratios of angle A in terms of sec A
Answer» We know that,sec A = 1/cos A⇒\xa0cos A = 1/sec Acos²A\xa0+ sin²A = 1⇒ sin²A = 1 – cos²A⇒ sin²A = 1 – (1/sec²A) (cosA= 1/secA)⇒ sin²A = (sec²A-1)/sec²A⇒sinA =\xa0√((sec²A-1)/sec²A)⇒sinA =\xa0√(sec²A-1) ÷ (secA)............................(i)sin A = 1/cosec A⇒ cosec A = 1/sin A⇒cosecA= secA\xa0÷√sec²A-1\xa0 (from eq i)Now,sec²A – tan²A = 1⇒ tan²A = sec²A\xa0+ 1⇒tanA = √sec²A\xa0+\xa01.....................................(ii)tan A = 1/cot A⇒ cot A = 1/tan A⇒cotA = 1/√sec²A\xa0+\xa01 (from eq ii)

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