Question

tan⁻¹2 +tan⁻¹3 is.....,Select Proper option from the given options.
(a) - π/4
(b) π/2
(c) 3π/4
(d) 3π/2

Answer 1

Answer:

option (a) is correct

Step-by-step explanation:

$tan^{-1}2+tan^{-1}3$

using, the formula

$\boxed{\bf\:tan^{-1}x+tan^{-1}y=tan^{-1}(\frac{x+y}{1-xy})}$

$=tan^{-1}(\frac{2+3}{1-(2)(3)})$

$=tan^{-1}(\frac{5}{1-6})$

$=tan^{-1}(\frac{5}{-5})$

$=tan^{-1}(-1)$

Using

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

$=-tan^{-1}(1)$

$=-\frac{\pi}{4}$

$\implies\:\boxed{\bf\:tan^{-1}2+tan^{-1}3=-\frac{\pi}{4}}$

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

$\sf{ { \tan}^{ - 1} 2 +{\tan}^{ - 1} 3 \: }$

FORMULA TO BE IMPLEMENTED

$\displaystyle \: \sf{ { \tan}^{ - 1} x +{\tan}^{ - 1} y = { \tan}^{ - 1} \bigg( \: \frac{x + y}{1 - xy} \bigg) \: \: \: \: \: where \: \: xy < 1 \: }$

CALCULATION

$\sf{ { \tan}^{ - 1} 2 +{\tan}^{ - 1} 3 \: }$

$\displaystyle = \sf{ \bigg( { \cot}^{ - 1} \frac{1}{2} +{\cot}^{ - 1} \frac{1}{3} \bigg) \: }$

$\displaystyle = \sf{ \bigg( \frac{\pi}{2} - { \tan}^{ - 1}\frac{1}{2} + \frac{\pi}{2} - {\tan}^{ - 1} \frac{1}{3} \bigg) \: }$

$\displaystyle = \sf{ \pi - \bigg( { \tan}^{ - 1} \frac{1}{2} +{\tan}^{ - 1} \frac{1}{3} \bigg) \: }$

$\displaystyle = \sf{ \pi - { \tan}^{ - 1} \bigg(\frac{ \frac{1}{2} + \frac{1}{3} }{1 - \frac{1}{2} \times \frac{1}{3} } \bigg) \: }$

$\displaystyle = \sf{ \pi - { \tan}^{ - 1} \bigg(\frac{ \frac{5}{ 6} }{1 - \frac{1}{6} } \bigg) \: }$

$\displaystyle = \sf{ \pi - { \tan}^{ - 1} \bigg(\frac{ \frac{5}{ 6} }{ \frac{5}{6} } \bigg) \: }$

$\displaystyle = \sf{ \pi - { \tan}^{ - 1} (1)\: }$

$\displaystyle = \sf{ \pi - \frac{\pi}{4} \: }$

$\displaystyle = \sf{ \frac{3\pi}{4} \: }$

RESULT

$\displaystyle \boxed {\sf{ \: \: \: { \tan}^{ - 1} 2 +{\tan}^{ - 1} 3 \: = \: \frac{3\pi}{4} \: \: \: }}$

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