trignometry challenge

GIVEN (a^2-b^2)sintheta +2abcostheta = a^2 +b^2

TO FIND = tantheta

i am not able to sole this problem need help...

#TrigonometricIdentities

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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Comments

how did u get it?

avn bha - 6 years, 10 months ago

$\tan \theta = \frac{a^{2}-b^{2}}{2ab}$

Rachit Jain - 6 years, 10 months ago

@avn bha , My answer is quite long. So please try to find any shorter way (if any) and do tell me... Dividing the given eqn. by $cos{ \theta }$ . You get the equation in terms of $\tan { \theta } \quad and\quad \sec { \theta }$ . Squaring both sides and converting $\sec ^{ 2 }{ \theta } =1+\tan ^{ 2 }{ \theta }$ and simplifying, u get an equation, quadratic in $\tan { \theta }$ which looks like

$4{ a }^{ 2 }{ b }^{ 2 }\tan ^{ 2 }{ \theta } -4ab({ a }^{ 2 }-{ b }^{ 2 })\tan { \theta } +{ ({ a }^{ 2 }-{ b }^{ 2 }) }^{ 2 }=0\\ Using\quad quadratic\quad formula\quad you\quad get\quad \\ \tan { \theta } =\frac { 4ab({ a }^{ 2 }-{ b }^{ 2 }) }{ 8{ a }^{ 2 }{ b }^{ 2 } } =\frac { { a }^{ 2 }-{ b }^{ 2 } }{ 2ab }$

CHEERS!!!:)

A Former Brilliant Member - 6 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$