Question

If sinx= k.sin.y, prove that, tan(x-y)/2=(k-1)/(k+1) tan(x+y)/2​

Answer 1

This question is just about a little manipulation skill, nothing very deep thinking stuff needed,so you really should ace such questions so that the other questions in your exam paper, probably tougher ones doesn't bother you much. So, let's try seeing if we can solve this or not.

sin (x) = k × sin (y)

We can bring the sin(y) to the LHS in the denominator,

sin (x)/sin (y) = k ----> (1)

Now,add 1 to both the sides of the equation (1)

[sin (x)/sin (y)] + 1 = k + 1

Do the LCM thing,

[sin (x) + sin (y)]/sin (y) = k + 1 ----> (2)

Next, subtract 1 from both the sides of the same equation i.e equation (1),

[sin (x)/sin (y)] -1 = k - 1

Again do the LCM thing,

sin (x) - sin (y)/sin (y) = k - 1 ----> (3)

Now if you want to proceed from here... you know you need to reduce the equation to as many less terms as possible,so equation (2) and (3) actually have the same denominator,so we can decide to divide the equation (2) by (3),the result :

[sin (x) + sin (y)]/[sin (x) - sin (y)] = [k + 1/ k -1]

You see the Nr and Dr, isn't it something you are familiar with,haan? Yeah, that's a direct formula,using the required formulae :

[2 sin (x+y/2) cos (x-y/2)]/ 2 [cos (x+y/2) sin (x-y/2)] = [k + 1/ k - 1]

Now next you need to just decide how you want to arrange your terms by looking at what you are actually asked to prove,so if you see we have sin and cos pair with same angles viz (x + y/2) and (x-y/2), so we at least know that we have arranged things for obtaining tan.

Club the same angles,

{2 [sin (x+y/2)]/2 [cos (x+y)/2)} × { cos (x - y/2)/ sin (x-y/2)} = k - 1/k + 1

Now if you observe,we actually have cos/sin which is cot, but we actually need tan,so we cross multiply,

[k - 1/k + 1] × tan (x+y/2) × cot (x-y/2) = 1

(k - 1/ k + 1) tan (x+y/2) = 1/cot (x-y/2)

Using the fact : 1/cot x = tanx,we get :

( k - 1/k + 1) tan(x+ y/2) = tan (x-y/2)

Proved :P

Well, you could have just used the C-D simply and skip the above 7-8 steps,it would have made the solution really short, you can try on your own ;)