Question

If, a sin x = b cos x = 2c tan x/1-tan^2x.
then proved that ...
(a²-b^2)^2 = 4c^2(a^2+ b^2).​

Answer 1

ANSWER:

Given that

${ \sf{asin \: x = bcos \: x = \dfrac{2ctan \: x}{1 - {tan}^{2}x } }}$

Taking the last term

${ \sf{ \frac{2c \: tanx}{1 - {tan}^{2} x} = c \times \frac{2tanx }{1 - {tan}^{2}x } = c \: tan2x }}$

It is in the form of

2tanx/1 - tan²x = tan2x

To prove

(a² - b²)² = 4c²(a² + b²)

asinx = bcosx

${ \sf{ \frac{a}{b} = \frac{cos \: x}{sin \: x} }}$

Squaring on both sides

${ \sf{ \dfrac{ {a}^{2} }{ {b}^{2} } = \dfrac{ {cos}^{2} x}{ {sin}^{2}x } }}$

Apply componendo & dividendo

${ \sf{ \dfrac{ {a}^{2} + {b}^{2} }{ {a}^{2} - {b}^{2} } = \dfrac{ {cos}^{2} x + {sin}^{2}x }{{cos}^{2}x - {sin}^{2} x } }}$

We know that

cos²x - sin²x = cos2x

cos²x + sin²x = 1

${ \sf{ \dfrac{ {a}^{2} + {b}^{2} }{ {a}^{2} - {b}^{2} } = \dfrac{1}{cos2x} }}$

Squaring on both sides

${ \sf{ {( \dfrac{{a}^{2} + {b}^{2} }{ {a}^{2} - {b}^{2} } )}^{2} = {( \dfrac{1}{cos2x}) }^{2} }}$

${ \sf{ \frac{ {( {a}^{2} - {b}^{2}) }^{2} }{ {( {a}^{2} - {b}^{2}) }^{2} } = \frac{ {sec}^{4}x. {b}^{2} {cos}^{2} x }{4 {c}^{2} {tan}^{2}x } }}$

Cross multiply

We get

${ \bf{( {a}^{2} - {b}^{2})^{2} = 4 {c}^{2}( {a}^{2} + {b}^{2} ) }}$

Hence proved