Find the value of sin (a+b) if tan a=4/3 and cot b= 5/12, 0^degrees<a<90^degrees and 0^degrees<b<90^degrees ?

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Answer 1 · 2018-05-01T09:00:49

#sin(a+b)=56/65#

Given, #tana=4/3 and cotb=5/12#

#rarrcota=3/4#

#rarrsina=1/csca=1/sqrt(1+cot^2a)=1/sqrt(1+(3/4)^2)=4/5#

#rarrcosa=sqrt(1-sin^2a)=sqrt(1-(4/5)^2)=3/5#

#rarrcotb=5/12#

#rarrsinb=1/cscb=1/sqrt(1+cot^2b)=1/sqrt(1+(5/12)^2)=12/13#

#rarrcosb=sqrt(1-sin^2b)=sqrt(1-(12/13)^2)=5/13#

Now, #sin(a+b)=sina*cosb+cosa*sinb#

#=(4/5)(5/13)+(3/5)*(12/13)=56/65#

Answer 2 · 2018-05-01T09:19:31

#sin(a+b)=56/65#

Here,

#0^circ < color(violet)( a ) < 90^circ=>I^(st)Quadrant=>color(blue)(All, fns. > 0.#

#0^circ < color(violet)(b) < 90^circ=>I^(st)Quadrant=>color(blue)(All,fns. > 0#

So,

#0^circ < color(violet)(a+b) < 180^circ=>I^(st) and II^(nd) Quadrant#

#=>color(blue)(sin(a+b) > 0#

Now,

#tana=4/3=>seca=+sqrt(1+tan^2a)=sqrt(1+16/9)=5/3#

#:.color(red)(cosa)=1/seca=color(red)(3/5#

#=>color(red)(sina)=+sqrt(1-cos^2a)=sqrt(1-9/25)=color(red)(4/5#

Also,

#cotb=5/12=>cscb=+sqrt(1+cot^2b)=sqrt(1+25/144)=13/12#

#:.color(red)(sinb)=1/cscb=color(red)(12/13#

#=>color(red)(cosb)=+sqrt(1-sin^2b)=sqrt(1-144/169)=color(red)(5/13#

Hence,

#sin(a+b)=sinacosb+cosasinb#

#=>sin(a+b)=4/5xx5/13+3/5xx12/13#

#sin(a+b)=20/65+36/65=56/65#