Mathematics — CBSE Class 10 board question

To prove: $\dfrac{1 + \cosec A}{\cosec A} = \dfrac{\cos^2 A}{1 - \sin A}$

LHS:

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

RHS:

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

Since LHS = RHS = $1 + \sin A$, the identity is proved. $\blacksquare$

(Used: $\dfrac{1}{\cosec A} = \sin A$ and $\cos^2 A = 1 - \sin^2 A$)

Source: Chapter 8, Section 8.4 Trigonometric Identities

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• Simplify both sides independently to the same expression — this is the standard method for "prove" questions.
• Key steps: $\frac{1}{\text{cosec }A} = \sin A$ on the LHS; factorising $\cos^2 A = (1-\sin A)(1+\sin A)$ on the RHS.
• Examiners award 1 mark each for correct simplification of LHS, correct factorisation on RHS, and concluding LHS = RHS.
• Don't cross-multiply; always work from one side or both sides separately.