Question #08ca1

2 Answers
P dilip_k
Feb 14, 2017

#LHS=(sec x sin x)/(tan x + cot x)#

#=(sec x sin x)/(sinx/cosx + cos x/sinx) #

#=(sec x sin x)/((sin^2x+cos^2x)/(cosxsinx)) #

#=secx*sinx*cosx*sinx#

#=sin^2x#

Jim G.
Feb 14, 2017

see explanation.

Explanation:

Making use of the following #color(blue)"trigonometric identities"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(tanx=sinx/cosx,cotx=cosx/sinx)color(white)(2/2)|)))#

#"and " color(red)(bar(ul(|color(white)(2/2)color(black)(secx=1/cosx)color(white)(2/2)|)))#

#"left side "=(secx xx sinx)/(tanx+cotx)#

#color(white)(xxxxxx)=(sinx/cosx)/(sinx/cosx+cosx/sinx)#

#color(white)(xxxxxx)=(sinx/cosx)/((sin^2x+cos^2x)/(sinxcosx))#

#color(white)(xxxxxx)=sinx/cancel(cosx) xx(sinxcancel(cosx))/1#

#color(white)(xxxxxx)=sin^2x="right side"rArr" verified"#

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