Question

If $y={\tan }^{-1}(secx+tanx)$ then $\frac{dy}{dx}=$

• A. $1$
• B. $\frac{1}{2}$
• C. $-1$
• D. $0$

Answer 1

Answer: (B)

$\frac{1}{2}$

Given $y=ta{n}^{-1}(secx+tanx)$

Differentiate on both sides w.r.t x

$\frac{dy}{dx}=\frac{1}{1+(secx+tanx{)}^2}\frac{d}{dx}(secx+tanx)$ $\because \frac{d}{dx}\left(ta{n}^{-1}\right(x)=\frac{1}{1+x^2})$

$=\frac{secxtanx+se{c}^{2}x}{1+se{c}^{2}x+ta{n}^{2}x+2secxtanx}$

$=\frac{secxtanx+se{c}^{2}x}{2se{c}^{2}x+2secxtanx}$ $\because (1+ta{n}^{2}x=se{c}^{2}x)$

$=\frac{secxtanx+se{c}^{2}x}{2[se{c}^{2}x+secxtanx]}$

$\therefore \frac{dy}{dx}=\frac{1}{2}$