Question

anyone in class 12th can help me in solving one inverse trigonometry question. (in attachment)​​

Answer 1

Answer:

$\frac{\pi }{4} + \frac{x}{2}$

Step-by-step explanation:

tan⁻¹($\frac{(1+sinx)}{cosx}$)

= tan⁻¹($\frac{sin^{2} \frac{x}{2}+cos^{2}\frac{x}{2}+2sin\frac{x}{2}cos\frac{x}{2}}{cos^2\frac{x}{2}-sin^2\frac{x}{2}}$)     [ ∵ sin²α+cos²α = 1 ,sin2α = 2sinαcosα, & cos2α=cos²α-sin²α ]

= tan⁻¹($\frac{(sin\frac{x}{2}+cos\frac{x}{2})^2} {(cos\frac{z}{2}+sin\frac{z}{2})(cos\frac{z}{2}-sin\frac{z}{2})}$)

= tan⁻¹($\frac{(sin\frac{x}{2}+cos\frac{x}{2})} {(cos\frac{z}{2}-sin\frac{z}{2})}$)

= tan⁻¹( $\frac{(sin\frac{x}{2}+cos\frac{x}{2})} {(cos\frac{z}{2}-sin\frac{z}{2})}$ ÷ $\frac{cos\frac{x}{2}}{cos\frac{x}{2}}$ )

= tan⁻¹($\frac{( \frac{sin\frac{x}{2}}{cos\frac{x}{2}} + \frac{cos\frac{x}{2}}{cos\frac{x}{2}} )} {( \frac{cos\frac{x}{2}}{cos\frac{x}{2}} - \frac{sin\frac{x}{2}}{cos\frac{x}{2}} )}$)

= tan⁻¹($\frac{1+tan\frac{x}{2}} {1-tan\frac{x}{2}}$)

= tan⁻¹($\frac{tan\frac{\pi}{4}+tan\frac{x}{2}} {1-tan\frac{\pi}{4}tan\frac{x}{2}}$)       [ ∵ tan$\frac{\pi}{4}$=1 ]

= tan⁻¹(tan$(\frac{\pi }{4} + \frac{x}{2} )$)           [ ∵ tan(α+β)= $\frac{tan\alpha +tan\beta}{1-tan\alpha tan\beta }$]

= $(\frac{\pi }{4} + \frac{x}{2} )$