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Multiple Choice Questions
1. tan(A/2) is equal to
(A): tanA/(1 + secA)
(B): 1/(cosecA + cotA)
(C): tanA/(1 + cosecA)
(D): 1/(secA + cotA)
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By: anil on 05 May 2019 02.17 am
Using double angle formula, we know that $$cos(2 heta) = cos^2 heta - sin^2 heta$$ => $$cos(2 heta) = (1 - sin^2 heta) - sin^2 heta$$ => $$cos(2 heta) = 1 - 2sin^2 heta$$ Replacing $$ heta$$ by $$frac{A}{2}$$, we get : => $$cos A = 1 - 2sin^2(frac{A}{2})$$ => $$2sin^2(frac{A}{2}) = 1 - cosA$$ => $$sin^2(frac{A}{2}) = frac{(1-cosA)}{2}$$ => $$sin(frac{A}{2}) = sqrt{frac{(1 - cos A)}{2}}$$ Similarly, => $$cos(frac{A}{2}) = sqrt{frac{(1 + cos A)}{2}}$$ Now, to find : $$tan(frac{A}{2})$$ = $$sin(frac{A}{2}) div cos(frac{A}{2})$$ = $$sqrt{frac{(1 - cos A)}{2}}$$ $$div$$ $$sqrt{frac{(1 + cos A)}{2}}$$ = $$sqrt{frac{(1 - cos A)}{2}}$$ $$ imes$$ $$sqrt{frac{2}{(1 + cos A)}}$$ = $$sqrt{frac{1-cosA}{1+cosA}}$$ = $$sqrt{frac{1-cosA}{1+cosA} imes frac{1+cosA}{1+cosA}}$$ = $$sqrt{frac{1-cos^2A}{(1+cosA)^2}} = sqrt{frac{sin^2A}{(1+cosA)^2}}$$ = $$frac{sinA}{1+cosA}$$ Dividing both numerator and denominator by $$(cosA)$$ = $$frac{tanA}{1+secA}$$ => Ans - (A)
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