Prove that: ^ -1 2ab a^2-b^2 + ^ -1 2xy x^2-y^2 = ^ -1 2 ^2- ^2 ,where =ax-by and =ay+bx.

^ -1 2ab a^2-b^2 + ^ -1 2xy x^2-y^2 = ^ -1 2 ^2- ^2 as =ax-by, =ay+bx L.H.S= ^ -1 2ab a^2-b^2 + ^ -1 2xy x^2-y^2 = ^ -1 2ab a^2-b^2 + 2xy x^2-y^2 1- ( 2ab a^2-b^2 ) ( 2xy x^2-y^2 ) Since, ^ -1 x+ ^ -1 y= ^ -1 x+y 1-xy = ^ -1 2abx^2-2aby^2+2xya^2-2xyb^2 (a^2-b^2 ) (x^2-y^2 ) a^2x^2-a^2y^2-b^2y^2+b^2y^2-4abxy (a^2-b^2 ) (x^2-y^2 ) = ^ -1 [ 2 (abx^2+xya^2-aby^2-xyb^2 ) a^2x^2+b^2y^2-2abxy-a^2y^2-b^2y^2-2abxy ] = ^ -1 [ 2 ax(bx+ay)-by(ay+bx) (ax-by)^2- (a^2y^2+b^2x^2+2abxy ) ] = ^ -1 [ 2(bx+ay)(ax-by) (ax-by)^2-(bx+ay)^2 ] = ^ -1 [2 ^2- ^2 ] Since, =ax-by, =ay+bx Hence, ^ -1 ( 2ab a^2-b^2 )+ ^ -1 ( 2xy x^2-y^2 )= ^ -1 (2 ^2- ^2 )

Prove that: ^ -1 2ab a^2-b^2 + ^ -1 2xy x^2-y^2 = ^ -1 2 ^2- ^2 ,…

Download App

Question

Prove that: $\tan^{-1}\frac{2\text{a}\text{b}}{\text{a}^2-\text{b}^2}+\tan^{-1}\frac{2\text{xy}}{\text{x}^2-\text{y}^2}=\tan^{-1}\frac{2\alpha\beta}{\alpha^2-\beta^2},$where $\alpha=\text{ax}-\text{by}$ and $\beta=\text{ay}+\text{bx}.$

✓

Answer

$\tan^{-1}\frac{2\text{a}\text{b}}{\text{a}^2-\text{b}^2}+\tan^{-1}\frac{2\text{xy}}{\text{x}^2-\text{y}^2}=\tan^{-1}\frac{2\alpha\beta}{\alpha^2-\beta^2}$ as $\alpha=\text{ax}-\text{by},\beta=\text{ay}+\text{bx}$
$\text{L.H.S}=\tan^{-1}\frac{2\text{a}\text{b}}{\text{a}^2-\text{b}^2}+\tan^{-1}\frac{2\text{xy}}{\text{x}^2-\text{y}^2}$
$=\tan^{-1}\begin{bmatrix}\frac{\frac{2\text{a}\text{b}}{\text{a}^2-\text{b}^2}+\frac{2\text{xy}}{\text{x}^2-\text{y}^2}}{1-\Big(\frac{2\text{ab}}{\text{a}^2-\text{b}^2}\Big)\Big(\frac{2\text{xy}}{\text{x}^2-\text{y}^2}\Big)}\end{bmatrix}$ $\Big\{\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\frac{\text{x}+\text{y}}{1-\text{xy}}\Big\}$
$=\tan^{-1}\begin{bmatrix}\frac{\frac{2\text{a}\text{b}\text{x}^2-2\text{a}\text{b}\text{y}^2+2\text{xy}\text{a}^2-2\text{xyb}^2}{\big(\text{a}^2-\text{b}^2\big)\big(\text{x}^2-\text{y}^2\big)}}{\frac{\text{a}^2\text{x}^2-\text{a}^2\text{y}^2-\text{b}^2\text{y}^2+\text{b}^2\text{y}^2-4\text{abxy}}{\big(\text{a}^2-\text{b}^2\big)\big(\text{x}^2-\text{y}^2\big)}}\end{bmatrix}$
$=\tan^{-1}\Bigg[\frac{2\big(\text{ab}\text{x}^2+\text{xy}\text{a}^2-\text{ab}\text{y}^2-\text{xy}\text{b}^2\big)}{\text{a}^2\text{x}^2+\text{b}^2\text{y}^2-2\text{abxy}-\text{a}^2\text{y}^2-\text{b}^2\text{y}^2-2\text{abxy}} \Bigg]$
$=\tan^{-1}\Bigg[\frac{2\{\text{ax}(\text{bx}+\text{ay})-\text{by}(\text{ay}+\text{bx})\}}{(\text{ax}-\text{by})^2-\big(\text{a}^2\text{y}^2+\text{b}^2\text{x}^2+2\text{abxy}\big)}\Bigg]$
$=\tan^{-1}\Bigg[\frac{2(\text{bx}+\text{ay})(\text{ax}-\text{by})\}}{(\text{ax}-\text{by})^2-(\text{bx}+\text{ay})^2}\Bigg]$
$=\tan^{-1}\Big[\frac{2\alpha\beta}{\alpha^2-\beta^2}\Big]$ $\{\text{Since},\alpha=\text{ax}-\text{by},\beta=\text{ay}+\text{bx}\}$
Hence,
$\tan^{-1}\bigg(\frac{2\text{a}\text{b}}{\text{a}^2-\text{b}^2}\bigg)+\tan^{-1}\bigg(\frac{2\text{xy}}{\text{x}^2-\text{y}^2}\bigg)=\tan^{-1}\bigg(\frac{2\alpha\beta}{\alpha^2-\beta^2}\bigg)$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from Inverse Trigonometric Functions→Inverse Trigonometric Functions — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Inverse Trigonometric Functions — Maths STD 12 Science — Question

Answer

Need a full question paper?

Explore more

Similar questions