Question

sin theta / cot theta+cosec theta = 2 + sin theta / cot theta - cosec theta prove that

Answer 1

[tex] \frac{sin\ \alpha }{cot\ \alpha\ +\ cosec\ \alpha } = \frac{sin\ \alpha [cosec\ \alpha\ -\ cot\ \alpha ]}{(cosec\ \alpha\ +\ cot\ \alpha )(cosec\ \alpha\ -\ cos\ \alpha) } \\ \\ = \frac{1 - cos \alpha }{1} \\ \\ Now,\ \ 2 + \frac{Sin\ \alpha }{ cot\ \alpha\ -\ cosec\ \alpha } = 2 + \frac{sin\ \alpha (cot\ \alpha + cosec\ \alpha )}{(cot\ \alpha -\ cosec \alpha )(cot\ \alpha +\ cosec\ \alpha )} \\ \\ 2 + \frac{[cos\ \alpha + 1]}{- 1} \\ 1 - cos\ \alpha \\ [/tex]

Answer 2

Answer:

Note : Here I am using A instead of theta.

LHS = sinA/(cotA+cosecA)

= sinA/[(cosA/sinA)+(1/sinA)]

= sinA/[(cosA+1)/sinA]

= sin²A/(1+cosA)

= (1-cos²A)/(1+cosA)

/* sin²A = 1 - cos²A */

= [(1+cosA)(1-cosA)]/(1+cosA)

/* a²-b² = (a+b)(a-b) */

= 1 - cosA ----(1)

RHS = 2+ [sinA/(cotA-cosecA)]

= 2+sinA/[(cosA/sinA)-(1/sinA)]

= 2+SinA/[(cosA-1)/sinA]

= 2+ [ sin²A/(cosA-1)]

= 2 - ( sin²A)/(1-cosA)

= 2- [ (1-cos²A)/(1-cosA)]

= 2-[(1+cosA)(1-cosA)/(1-cosA)]

= 2 - ( 1+cosA)

= 2-1-cosA

= 1-cosA ----(2)

Form (1) & (2) , we conclude that,

(1) = (2)

LHS = RHS

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