CBSE Class 12 Mathematics 2023: question 27(b)

Hints

Open one hint at a time and try again before the next.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. $y=\tan x+\sec x=\frac{\sin x+1}{\cos x}$

   By the quotient rule:

   $\frac{dy}{dx}=\frac{\cos x\cdot \cos x-(1+\sin x)(-\sin x)}{{\cos }^{2}x}=\frac{{\cos }^{2}x+{\sin }^{2}x+\sin x}{{\cos }^{2}x}$

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

   1½ marks

2. Differentiate again:

   $\frac{d^{2}y}{d{x}^2}=\frac{(1-\sin x)\cdot 0-1\cdot (-\cos x)}{(1-\sin x{)}^2}=\frac{\cos x}{(1-\sin x{)}^2}$

   Hence proved.

   1½ marks

Answer: $\frac{d^{2}y}{d{x}^2}=\frac{\cos x}{(1-\sin x{)}^2}$ (proved)

Revise the chapter

• Chapter: Continuity and Differentiability (notes, formulas, where marks are lost and every past question)
• Topic: Parametric and Second Order Derivatives