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Multiple Choice Questions
1. (secA - tanA)/(cosecA + cotA) is equal to
(A): (secA - cotA) / (cosecA + tanA)
(B): (secA - cosA) / (cosecA + sinA)
(C): (cosecA - cotA) / (secA + tanA)
(D): (secA + cosA) / (cosecA - sinA)
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By: anil on 05 May 2019 02.16 am
Expression : (secA - tanA)/(cosecA + cotA) = $$frac{(secA-tanA)}{(cosecA+cotA)}$$ Multiplying both numerator and denominator by $$(cosecA-cotA)$$ = $$frac{(secA-tanA)}{(cosecA+cotA)}$$ $$ imes frac{(cosecA-cotA)}{(cosecA-cotA)}$$
= $$frac{(secA-tanA)(cosecA-cotA)}{cosec^2A-cot^2A}$$ = $$(secA-tanA)(cosecA-cotA)$$ Multiplying both numerator and denominator by $$(secA+tanA)$$
= $$(secA-tanA)(cosecA-cotA)$$ $$ imes frac{(secA+tanA)}{(secA+tanA)}$$
= $$frac{(cosecA-cotA)(sec^2A-tan^2A)}{(secA+tanA)}$$ = $$frac{(cosecA-cotA)}{(secA+tanA)}$$ => Ans - (C)
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SSC CHSL 16 Jan 2017 Afternoon Shift
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= $$frac{(secA-tanA)(cosecA-cotA)}{cosec^2A-cot^2A}$$ = $$(secA-tanA)(cosecA-cotA)$$ Multiplying both numerator and denominator by $$(secA+tanA)$$
= $$(secA-tanA)(cosecA-cotA)$$ $$ imes frac{(secA+tanA)}{(secA+tanA)}$$
= $$frac{(cosecA-cotA)(sec^2A-tan^2A)}{(secA+tanA)}$$ = $$frac{(cosecA-cotA)}{(secA+tanA)}$$ => Ans - (C)