Question

The value of sec[tan⁻¹(b+a/b-a)- tan⁻¹(a/b)] is......,Select Proper option from the given options.
(a) 1
(b) √2
(c) 2
(d) 4

Answer 1

Answer:

$\sqrt{2}$

Step-by-step explanation :

In this question

sec[$tan^{-1}{\frac{b+a}{b-a}} - tan^{-1}{\frac{a}{b}$]

property $tan^{-1}{x} - tan^{-1}{y}$ = $tan^{-1}\frac{x-y}{1+xy}$

We have been given that

x = $\frac{b+a}{b-a}$ and y = $\frac{a}{b}$

so we put the value of x and y,

$\implies$ sec[$tan^{-1}{\frac{b+a}{b-a}} - tan^{-1}{\frac{a}{b}$}] = $sec[tan^{-1}\frac{\frac{b+a}{b-a}-\frac{a}{b}}{1+(\frac{b+a}{b-a})\frac{a}{b}}]$

$\implies$ $sec[tan^{-1}\frac{\frac{b^2+ab-ab+a^2}{b^2-ab}}{\frac{b^2-ab+ab+a^2}{b^2-ab}}]$

$\implies[tex] [tex]sec[tan^{-1}(1)]$

Therefore, $tan^{-1}(1)$ = 45°

So, sec45° = √2

Hence, value of sec[$tan^{-1}{\frac{b+a}{b-a}} - tan^{-1}{\frac{a}{b}$}] = sec45° = √2

So, option (b) is right.

Answer 2

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

TO DETERMINE

$\displaystyle \sf{ \sec \bigg [ { \tan}^{ - 1} \bigg( \frac{b + a}{b - a} \bigg) - { \tan}^{ - 1} \bigg( \frac{a}{b} \bigg) \bigg] \: \: }$

CALCULATION

NOW

$\displaystyle \sf{ \bigg [ { \tan}^{ - 1} \bigg( \frac{b + a}{b - a} \bigg) - { \tan}^{ - 1} \bigg( \frac{a}{b} \bigg) \bigg] \: \: }$

$= \displaystyle \sf{ \bigg [ { \tan}^{ - 1} \bigg( \frac{1 + \frac{a}{b} }{1 - \frac{a}{b} } \bigg) - { \tan}^{ - 1} \bigg( \frac{a}{b} \bigg) \bigg] \: \: }$

$\displaystyle \: \: \: \sf{Let \: \: \frac{a}{b} = \tan \theta } \: \: \: \: so \: that \: \: \: \theta = { \tan}^{ - 1} \bigg(\frac{a}{b} \bigg)$

So

$\displaystyle \sf{ \bigg [ { \tan}^{ - 1} \bigg( \frac{b + a}{b - a} \bigg) - { \tan}^{ - 1} \bigg( \frac{a}{b} \bigg) \bigg] \: \: }$

$= \displaystyle \sf{ \bigg [ { \tan}^{ - 1} \bigg( \frac{1 + \frac{a}{b} }{1 - \frac{a}{b} } \bigg) - { \tan}^{ - 1} \bigg( \frac{a}{b} \bigg) \bigg] \: \: }$

$= \displaystyle \sf{ \bigg [ { \tan}^{ - 1} \bigg( \frac{1 + \tan \theta}{1 - \tan \theta} \bigg) - { \tan}^{ - 1} \bigg( \tan \theta\bigg) \bigg] \: \: }$

$= \displaystyle \sf{ \bigg [ { \tan}^{ - 1} \bigg( \frac{ \tan \frac{\pi}{4} + \tan \theta}{1 - \tan \frac{\pi}{4} \tan \theta} \bigg) - { \tan}^{ - 1} \bigg( \tan \theta\bigg) \bigg] \: \: }$

$= \displaystyle \sf{ \bigg [ { \tan}^{ - 1} \tan\bigg( \frac{\pi}{4} + \theta \bigg) - { \tan}^{ - 1} \bigg( \tan \theta\bigg) \bigg] \: \: }$

$= \displaystyle \sf{ \bigg [\bigg( \frac{\pi}{4} + \theta \bigg) - \theta\bigg] \: \: }$

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

RESULT

$\displaystyle \sf{ \sec \bigg [ { \tan}^{ - 1} \bigg( \frac{b + a}{b - a} \bigg) - { \tan}^{ - 1} \bigg( \frac{a}{b} \bigg) \bigg] \: \: }$

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

$= \sf{ \sqrt{2} \: }$

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