Question #2144a

2 Answers

It is proved below:

Explanation:

#tan a+sec a=x/y#
#or,sin a/cos a+1/cos a=x/y#
#or,(sin a+1)/cos a=x/y#
#or,(sin a+1)^2/cos^2 a=x^2/y^2#
#or,(sina+1)^2/(1-sin^2a)=x^2/y^2#
#or,(sina+1)^2/((1+sin a)(1-sina))=x^2/y^2#
#or,(sina+1)/(1-sina)=x^2/y^2#
#or,(sina+1-1+sina)/(sina+1+1-sina)=(x^2-y^2)/(x^2+y^2)" "#[By division-addition method]
#or,(2sina)/(2)=(x^2-y^2)/(x^2+y^2)#
#so,sina=(x^2-y^2)/(x^2+y^2)color(blue)([Proved.])#

Dec 18, 2017

Given # sec(a)+tan(a)=x/y......(1)#

We know

#sec^2(a)-tan^2(a)=1#

#=>(sec(a)+tan(a))(sec(a)-tan(a))=1#

#=>sec(a)-sec(a)=1/(sec(a)+tan(a)) #

#=>sec(a)-tan(a)=1/(x/y) #

#=>sec(a)-tan(a)=y/x.......(2) #

Adding (1) and (2) we get

#=>2sec(a)=x/y+y/x=(x^2+y^2)/(xy) #

#=>sec(a)=(x^2+y^2)/(2xy)....(3) #

Now subtracting (2) from (1) we get

#=>2tan(a)=x/y-y/x=(x^2-y^2)/(xy) #

#=>tan(a)=(x^2-y^2)/(2xy) .....(4)#

Dividing (4) by (3) we have

#tan(a)/sec(a)=(x^2-y^2)/(x^2+y^2)#

#=>sin(a)=(x^2-y^2)/(x^2+y^2)#