Expand #sec(alpha+beta)sec(alpha-beta)# in terms of trigonometric ratios of #alpha# and #beta#?

1 Answer
Shwetank Mauria
Dec 8, 2017

#sec(alpha+beta)sec(alpha-beta)=1/(cos^2alpha-sin^2beta)#

Explanation:

#sec(alpha+beta)sec(alpha-beta)#

= #1/(cos(alpha+beta)cos(alpha-beta))#

= #1/((cosalphacosbeta-sinalphasinbeta)(cosalphacosbeta+sinalphasinbeta))#

= #1/(cos^2alphacos^2beta-sin^2alphasin^2beta)#

= #1/(cos^2alpha(1-sin^2beta)-(1-cos^2alpha)sin^2beta)#

= #1/(cos^2alpha-cos^2alphasin^2beta-sin^2beta+cos^2alphasin^2beta)#

= #1/(cos^2alpha-sin^2beta)#

Related questions
Impact of this question
2234 views around the world
You can reuse this answer
Creative Commons License