Question #08ca1

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Question #08ca1

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Answer 1 · 2017-02-14T14:54:39

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

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

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

$=secx*sinx*cosx*sinx$

$=sin^2x$

Answer 2 · 2017-02-14T15:01:28

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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Question #08ca1

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