Question

if tan a = n tan b and sin a = m sin b prove that cos^2 a = (m^2 - 1)/(n^2 - 1).

Answer 1

So clearly in the question we have to eleminate angle b first,

and with tan and sin , we can use

cosec^2B - cot^2B = 1


So let's turn them into our required forms first,


$\tan(a) = n \tan(b) \\ \\ \frac{1}{ \tan(b) } = \frac{n}{ \tan(a) } \\ \\ \cot(b) = \frac{n}{ \tan(a) }$


Similarly, with sin

$\csc(b ) = \frac{m}{ \sin(a) }$


Square both sides of cot and cosec/csc


${ \cot }^{2} b= ({ \frac{n}{ \tan(a) } })^{2} = \frac{ {n}^{2} }{ { \tan}^{2}a }$
-----(i)


Similarly,
${ \csc }^{2}b = \frac{ {m}^{2} }{ { \sin }^{2}a }$
-------(ii)

Subtracting (i) from (ii)

${ { \csc}^{2}b } - { \cot }^{2} b = \frac{ {m}^{2} }{ { \sin }^{2}a } - \frac{ {n}^{ 2 } }{{ \tan }^{2}a} \\ \\ \frac{ {m}^{2} }{ { \sin }^{2}a } - \frac{ {n}^{ 2 } }{{ \tan }^{2}a} = 1 \\ \\ \\ \\ \frac{ {m}^{2} { \tan}^{2}a - {n}^{2} { \sin }^{2}a }{ { \sin }^{2}a \: \: { \tan }^{2}a } = 1 \\ \\ {m}^{2} - {n}^{2} { \cos }^{2} a = { \sin }^{2}a \\ \\ {m}^{2} - {n}^{2} { \cos }^{2} a =1 - { \cos }^{2} a \\ \\ {m}^{2} - 1 = { \cos}^{2}a( {n}^{2} - 1) \\ \\ { \cos }^{2} a = \frac{ {m}^{2} - 1 }{ {n}^{2 } - 1 }$



Hence proved.