Prove that (cos A - sin A + 1) / (cos A + sin A

Q) Prove that $\frac{(cos A - sin A + 1)}{(cos A + sin A - 1)}$ = cosec A + cot A

Q29 B– Sample Question Paper – Set 1 – Maths Standard – CBSE 2026

Ans:

Let’s start from LHS:

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

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

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

Let’s consider: cos A = P and ( 1 – sin A) = Q

∴ LHS = $\frac{(P + Q)}{(P - Q)}$

To normalize the denominator, let’s multiple numerator and denominator both by denominator’s conjugate i.e. (P + Q):

∴ LHS = $\frac{(P + Q)}{(P - Q)} \times \frac{(P + Q)}{(P + Q)}$

= $\frac{(P + Q)(P + Q)}{(P - Q)(P + Q)}$

= $\frac{(P^2 + Q^2 + 2 P Q) }{ (P^2 - Q^2)}$

Let’s substitute values of P and Q, we get:

= $\frac{[(cos A)^2 + ( 1 - sin A)^2 + 2 (cos A)( 1 - sin A)]}{ [(cos A)^2 - ( 1 - sin A)^2] }$

= $\frac{ [(cos^2 A + ( 1 + sin^2 A - 2 sin A) + 2 cos A - 2 cos A sin A)] }{ [cos^2 A - ( 1 + sin^2 A - 2 sin A)] }$

= $\frac{ [cos^2 A + 1 + sin^2 A - 2 sin A + 2 cos A - 2 cos A sin A] }{ [cos^2 A - 1 - sin^2 A + 2 sin A)] }$

= $\frac{ [cos^2 A + sin^2 A + 1 - 2 sin A + 2 cos A - 2 cos A sin A] }{ [cos^2 A - 1 - sin^2 A + 2 sin A)] }$

Since $cos^2A + sin ^2 A = 1$

= $\frac{ [(1) + 1 - 2 sin A + 2 cos A - 2 cos A sin A] }{ [cos^2 A - (cos^2 A + sin^2 A) - sin^2 A + 2 sin A] }$

= $\frac{ [2 - 2 sin A + 2 cos A - 2 cos A sin A] }{ [\cancel{cos^2 A} - \cancel{cos^2 A} - sin^2 A - sin^2 A + 2 sin A] }$

= $\frac{ [2 - 2 sin A + 2 cos A - 2 cos A sin A] }{ [- 2 sin^2 A + 2 sin A] }$

Dividing whole expression by 2, we get:

LHS = $\frac{ [1 - sin A + cos A - cos A sin A] }{ [- sin^2 A + sin A] }$

= $\frac{ [(1 - sin A) + cos A (1 - sin A)] }{ [sin A - sin ^2 A)] }$

= $\frac{ [(1 - sin A)(1 + cos A)] }{ [sin A (1- sin A)]}$

= $\frac{ [\cancel{ (1 - sin A)}(1 + cos A)] }{ [sin A \cancel{ (1 - sin A)}]}$

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

= $\frac{1}{sin A} + \frac{cos A}{sin A}$

= cosec A + cot A = RHS

Hence Proved!

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