Question

2cot⁻¹ 2 + cosec⁻¹ 5/3 =π/2,Prove it.

Answer 1

$\underline{\textbf{To prove:}}$

$\mathsf{2\;cot^{-1}(2)+cosec^{-1}\left(\dfrac{5}{3}\right)=\dfrac{\pi}{2}}$

$\underline{\textbf{Solution:}}$

$\underline{\textsf{Formula used:}}$

$\mathsf{1.\;\;2\;tan^{-1}x=tan^{-1}\left(\dfrac{2x}{1-x^2}\right)}$

$\mathsf{2.\;\;sec^{-1}x+cosec^{-1}x=\dfrac{\pi}{2}}$

$\mathsf{3.\;\;cot^{-1}x=tan^{-1}\left(\dfrac{1}{x}\right)}$

$\mathsf{Consider}$

$\mathsf{2\;cot^{-1}(2)}$

$\mathsf{=2\;tan^{-1}\left(\dfrac{1}{2}\right)}$

$\textsf{Using formula (1),}$

$\mathsf{=tan^{-1}\left(\dfrac{2\left(\dfrac{1}{2}\right)}{1-\left(\dfrac{1}{2}\right)^2}\right)}$

$\mathsf{=tan^{-1}\left(\dfrac{1}{1-\dfrac{1}{4}}\right)}$

$\mathsf{=tan^{-1}\left(\dfrac{1}{\dfrac{3}{4}}\right)}$

$\mathsf{=tan^{-1}\left(\dfrac{4}{3}\right)}$

$\mathsf{Take,\;A=tan^{-1}\left(\dfrac{4}{3}\right)}$

$\mathsf{tanA=\dfrac{4}{3}}$

$\mathsf{sec^2A=1+tan^2A}$

$\mathsf{sec^2A=1+\dfrac{16}{9}}$

$\mathsf{sec^2A=\dfrac{25}{9}}$

$\mathsf{secA=\sqrt{\dfrac{25}{9}}}$

$\mathsf{secA=\dfrac{5}{3}}$

$\mathsf{A=sec^{-1}\left(\dfrac{5}{3}\right)}$

$\therefore\mathsf{Now,}$

$\mathsf{2\;cot^{-1}(2)+cosec^{-1}\left(\dfrac{5}{3}\right)}$

$\mathsf{=sec^{-1}\left(\dfrac{5}{3}\right)+cosec^{-1}\left(\dfrac{5}{3}\right)}$

$\textsf{Using formula (2)}$

$\mathsf{=\dfrac{\pi}{2}}$

$\implies\boxed{\mathsf{2\;cot^{-1}(2)+cosec^{-1}\left(\dfrac{5}{3}\right)=\dfrac{\pi}{2}}}$

$\underline{\textbf{Find more:}}$

Prove that: tan inverse 63/16=sin inverse 5/13+cos inverse 3/5

https://brainly.in/question/11002993

Answer 2

SOLUTION

TO PROVE

$\displaystyle \sf{2 { \cot}^{ - 1} 2 + { \csc}^{ - 1} \frac{5}{3} = \frac{\pi}{2} }$

PROOF

LHS

$\displaystyle \sf{ = 2 { \cot}^{ - 1} 2 + { \csc}^{ - 1} \frac{5}{3} }$

$\displaystyle \sf{ = 2 \: { \tan}^{ - 1} \frac{1}{2} + { \csc}^{ - 1} \frac{5}{3} }$

$\displaystyle \sf{ = { \tan}^{ - 1} \frac{2. \frac{1}{2} }{1 - { (\frac{1}{2} )}^{2} } + { \csc}^{ - 1} \frac{5}{3} }$

$\displaystyle \sf{ = { \tan}^{ - 1} \frac{1 }{1 - \frac{1}{4} } + { \csc}^{ - 1} \frac{5}{3} }$

$\displaystyle \sf{ = { \tan}^{ - 1} \frac{1 }{ \frac{4 - 1}{4} } + { \csc}^{ - 1} \frac{5}{3} }$

$\displaystyle \sf{ = { \tan}^{ - 1} \frac{1 }{ \frac{3}{4} } + { \csc}^{ - 1} \frac{5}{3} }$

$\displaystyle \sf{ = { \tan}^{ - 1} \frac{4 }{3 } + { \csc}^{ - 1} \frac{5}{3} }$

$\displaystyle \sf{ = { \sec}^{ - 1} \sqrt{1 + { \bigg(\frac{4}{3} \bigg)}^{2} } + { \csc}^{ - 1} \frac{5}{3} }$

$\displaystyle \sf{ = { \sec}^{ - 1} \sqrt{1 + { \frac{16}{9} }} + { \csc}^{ - 1} \frac{5}{3} }$

$\displaystyle \sf{ = { \sec}^{ - 1} \sqrt{ { \frac{9 + 16}{9} }} + { \csc}^{ - 1} \frac{5}{3} }$

$\displaystyle \sf{ = { \sec}^{ - 1} \sqrt{ { \frac{25}{9} }} + { \csc}^{ - 1} \frac{5}{3} }$

$\displaystyle \sf{ = { \sec}^{ - 1} \frac{5}{3} + { \csc}^{ - 1} \frac{5}{3} }$

$\displaystyle \sf{ = \frac{\pi}{2} }$

= RHS

Hence proved

Note : In the above proof csc refers to cosec

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. Cos^-1(1-9^x/1+9^x) differentiate with respect to x

https://brainly.in/question/17839008

2. prove that the curl of the gradient of

(scalar function) is zero

https://brainly.in/question/19412715