Question

If α=cos⁻¹(4/5),β=tan⁻¹(2/3),α,β ∈(0,π/2),then α-β=.......,Select Proper option from the given options.
(a) sin⁻¹2/√13
(b) tan⁻¹(1/18)
(c) cos⁻¹(1/5√3)
(d) sin⁻¹(6/5√13)

Answer 1

Dear Student,

Answer:cos⁻¹(18/5√13)

sin⁻¹(1/5√13)

tan⁻¹(1/6)

Solution:


$\alpha = {cos}^{ - 1} \frac{4}{5} \\ \\ \beta = { \tan }^{ - 1} \frac{2}{3} \\ \\ \alpha - \beta = {cos}^{ - 1} \frac{4}{5} - { \tan }^{ - 1} \frac{2}{3} \\ since \: \: \: {tan}^{ - 1} = {cos}^{ - 1} ( \frac{1}{ \sqrt{1 + {x}^{2} } } ) \\ \\ \alpha - \beta = {cos}^{ - 1} \frac{4}{5} - {cos}^{ - 1} ( \frac{1}{ \sqrt{1 + \frac{4}{9} } } \\ \\ = {cos}^{ - 1} \frac{4}{5} - {cos}^{ - 1} ( \frac{1}{ \sqrt{ \frac{9 + 4}{9} } } )\\ = {cos}^{ - 1} \frac{4}{5} - {cos}^{ - 1} ( \frac{3}{ \sqrt{ 13 } } )\\ \\ {cos}^{ - 1} x - {cos}^{ - 1} y = {cos}^{ - 1} (xy + \sqrt{1 - {x}^{2} } ) \sqrt{1 - {y}^{2} } ) \\ \\ = {cos}^{ - 1} ( \frac{12}{5 \sqrt{13} } + \sqrt{1 - { (\frac{4}{5} )}^{2} } ) \sqrt{1 - {( \frac{3}{ \sqrt{13}) } }^{2} } ) \\ \\ = {cos}^{ - 1} ( \frac{12}{5 \sqrt{13} } + \frac{6}{5 \sqrt{13} } ) \\ \\ \alpha - \beta = {cos}^{ - 1} ( \frac{18}{5 \sqrt{13} } )$
there might some mistakes in given options.

if we convert cos inverse in terms of sin inverse
ie
${cos}^{ - 1} x = {sin}^{ - 1} \sqrt{1 - {x}^{2} } \\ {cos}^{ - 1} ( \frac{18}{5 \sqrt{13} } = {sin}^{ - 1} \sqrt{1 - \frac{324}{325} } \\ \\ \\ \alpha - \beta = {sin}^{ - 1} ( \frac{1}{5 \sqrt{13} } )$
or in terms of tan answer would be tan⁻¹(1/6)

hope it helps you