Question #2144a

2 Answers
Dec 18, 2017

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)