Simplify #cosx+sinx/cotx#?
2 Answers
Dec 15, 2017
Please see below.
Explanation:
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Dec 15, 2017
Explanation:
#"using the "color(blue)"trigonometric identities"#
#•color(white)(x)cotx=cosx/sinx#
#•color(white)(x)secx=1/cosx#
#•color(white)(x)cos^2x+sin^2x=1#
#"consider the left hand side"#
#cosx+sinx/cotx#
#=cosx+sinx/(cosx/sinx)#
#=cosx+sinx/1xxsinx/cosx#
#=(cos^2x+sin^2x)/cosxlarrcolor(blue)"LCD of cosx"#
#=1/cosx#
#=secx=" right hand side "rArr"verified"#