Question

solve thus inverse trigonometric function tell me answer ​

Answer 1

Answer:

t

Step-by-step explanation:

tan^-1(x) – tan^-1(y)

=tan^-1 {(x-y)/(1+xy)}

tan^-1(x/y) – tan^-1 (x-y)/(x+y)

=tan^-1 {(x/y)-(x-y)/(x+y)} / {1+ x(x-y)/y(x+y)}

=tan^-1 {(x²+xy-xy+y²)/(xy+y²-xy+y²)}

=tan^-1 (1)

=π/4 ....(c)

Answer 2

Substitute,

• $x=\sin\theta$

• $y=\cos\theta$

where $\theta\in\left(\-\dfrac{\pi}{2},\ \dfrac{\pi}{2}\right).$

Then,

$\longrightarrow \tan^{-1}\left(\dfrac{x}{y}\right)-\tan^{-1}\left(\dfrac{x-y}{x+y}\right)=\tan^{-1}\left(\dfrac{\sin\theta}{\cos\theta}\right)-\tan^{-1}\left(\dfrac{\sin\theta-\cos\theta}{\sin\theta+\cos\theta}\right)$

$\longrightarrow \tan^{-1}\left(\dfrac{x}{y}\right)-\tan^{-1}\left(\dfrac{x-y}{x+y}\right)=\tan^{-1}\left(\tan\theta\right)-\tan^{-1}\left(\dfrac{\tan\theta-1}{\tan\theta+1}\right)$

Since $\dfrac{1-\tan\theta}{1+\tan\theta}=\tan\left(\dfrac{\pi}{4}-\theta\right),$

$\longrightarrow \tan^{-1}\left(\dfrac{x}{y}\right)-\tan^{-1}\left(\dfrac{x-y}{x+y}\right)=\theta-\tan^{-1}\left(-\tan\left(\dfrac{\pi}{4}-\theta\right)\right)$

Since $\tan^{-1}(-x)=-\tan^{-1}x,$

$\longrightarrow \tan^{-1}\left(\dfrac{x}{y}\right)-\tan^{-1}\left(\dfrac{x-y}{x+y}\right)=\theta+\tan^{-1}\left(\tan\left(\dfrac{\pi}{4}-\theta\right)\right)$

$\longrightarrow \tan^{-1}\left(\dfrac{x}{y}\right)-\tan^{-1}\left(\dfrac{x-y}{x+y}\right)=\theta+\dfrac{\pi}{4}-\theta$

$\longrightarrow\underline{\underline{\tan^{-1}\left(\dfrac{x}{y}\right)-\tan^{-1}\left(\dfrac{x-y}{x+y}\right)=\dfrac{\pi}{4}}}$

Hence (C) is the answer.