Question

Solve This TRIGONOMERIY Question. 40 "#POINTS". PLZ Answer Quickly In need your help

Answer 1

Answer
............




Tan A = n tan B can be written as,
======
{tanA} ===
_____________
{tanB} = n



================

tanA *
___________
{1}{tanB}. = n

_______________
{sinA}. cosA} *
______________
{cosB}{sinB} = n

{sinAcosB}{
cosAsinB} = n




A = m sin B ----

Substitute (2) in (1), we get

{m sinB cosB}
--------. )
= {cosA sinB} = n

{m cosB}{cosA} =n

m cos B = n cosA


On squaring both sides, we get

n^2 cos^2A = m^2 cos^2B

= m^2(1 - sin^2B)

Equation can be written as sinB = sinA/m

= m^2(1 -
________________________
{sin^2A}{m^2} )



------------

m^2 - sin^2A

m^2 - 1 + cos^2A


n^2cos^2A - cos^2A = m^2 - 1

cos^2A = \frac{m^2 - 1}{n^2 - 1}


LHS = RHS.

Answer 2

Hey friends ..

We have to eliminate B. So ,we find cosecB and CotB from the given relations and use the identity cosec²CB -cot²B=1 .


SinA=msinB=>cosecB=m/sinA------------1)


tanA=ntanB=>cotB=n/tanA-------2)


squaring 1 ) and 2} and subtracting we get

m²/sin²A-n²/tan²A=cosec²B-cot²B

As we know cosec²B-cot²B=1


m²/sin²A-n²/sin²A/cos²A=1

=>m²-n²cos²A=>sin²A

=>m²-n²cos²A=1-cos²A

=>(n²-1)cos²A=m²-1

=>cos²A=m²-1/n²+1 prooved here...

Hope it helps you..

@Rajukumar111