Inverse trig. question (1 Viewer)

[lsdpoon1337]

Show that:

tan-1(4) - tan-1(3/5)= pi/4,

thanks

 

[lyounamu]

Show that:

tan-1(4) - tan-1(3/5)= pi/4,

thanks

sin-1(sin (tan-1(4) - tan-1(3/5))) =

Let p = (tan-1(4))
tan p = 4 (draw a triangle here)

Let q = (tan-1(3/5))
tan q = 3/5 (draw a triangle here)

So sin-1(sin (tan-1(4) - tan-1(3/5))) = sin-1( sin (p-q)) = sin-1 (sinpcosq - cospsinq)
= sin-1(4/SQRT(17) . 5/SQRT(34) - 1/SQRT(17) . 3/SQRT(34))
= sin-1(20/17SQRT(2) - 3/SQRT(2))
= sin -1(17/17SQRT(2))
= sin-1(1/SQRT(2))
= pi/4
proved/

 

[munch0r]

sin-1(sin (tan-1(4) - tan-1(3/5))) =

Let p = (tan-1(4))
tan p = 4 (draw a triangle here)

Let q = (tan-1(3/5))
tan q = 3/5 (draw a triangle here)

So sin-1(sin (tan-1(4) - tan-1(3/5))) = sin-1( sin (p-q)) = sin-1 (sinpcosq - cospsinq)
= sin-1(4/SQRT(17) . 5/SQRT(34) - 1/SQRT(17) . 3/SQRT(34))
= sin-1(20/17SQRT(2) - 3/SQRT(2))
= sin -1(17/17SQRT(2))
= sin-1(1/SQRT(2))
= pi/4
proved/

usually the only thing you have to do is, take tan of both sides and use double angle formula for LHS....

 

[lyounamu]

usually the only thing you have to do is, take tan of both sides and use double angle formula for LHS....

Can you share your working-out?

 

[munch0r]

Can you share your working-out?

tan-1(4) - tan-1(3/5)= pi/4
tan (tan-1(4) - tan-1(3/5)) = tan (pi/4)
RHS = 1
LHS (using double angle formula) = (4 - 3/5) / (1 + 12/5) = 1
LHS = RHS

 

[lyounamu]

tan-1(4) - tan-1(3/5)= pi/4
tan (tan-1(4) - tan-1(3/5)) = tan (pi/4)
RHS = 1
LHS (using double angle formula) = (4 - 3/5) / (1 + 12/5) = 1
LHS = RHS

Nah, you cannot do that.

It's a "show" question which means you have to work from left to right, not from both sides.

if it was a "prove" question, your working-out is acceptable.

 

[lsdpoon1337]

Thanks, you have given me light !!!