Question

cosA-sinA+1 / cosA +sinA-1 =cosecA + cotA
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Answer 1

Heya friend !

L.H.S

⇒ $\frac{cos A - sin A + 1}{cos A + sin A - 1}$

After rationalizing, we get,

⇒ $\frac{(cos A - sin A +1)(cos A + sin A + 1)}{(cos A + sin A - 1)(cos A + sin A + 1)}$

Now, we can put the identity of :-

$(a-b)(a+b) = a^{2} - b^{2}$  in both numerator and denominator.

In numerator,

a = cos A + 1                    b = sin A

In denominator,

a = cos A + sin A             b = 1

For further clarification, we can rewrite the equation as :-

⇒ $\frac{[(cos A + 1) - sin A] * [(cos A + 1)+sinA]}{[(cos A + sin A)-1]*[(cos A + sin A)+1]}$  

⇒ $\frac{(cosA+1)^{2}-sin^{2} A }{(cosA+sinA)^{2} - 1^{2} }$

⇒ $\frac{cos^{2}A + 1 + 2cos A - sin^{2}A }{cos^{2}A +sin^{2}A + 2sinAcosA - 1 }$

⇒ $\frac{2cos^{2}A+2cosA }{2sinAcosA}$                                     [$cos^{2}A = 1 - sin^{2}A$] and [$sin^{2}A+ cos^{2}A =1$]

⇒ $\frac{2cosA(cosA+1)}{2cosAsinA}$

⇒  $\frac{cosA+1}{sinA}$

By taking the denominator separately with both the terms in the numerator, we get,            

⇒ $\frac{cosA}{sinA} + \frac{1}{sinA}$

⇒ $cotA + cosec A$            = R.H.S.

Hence Proved !                        

Thanks !

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